Same, I just do. Eny meany myny mo( don’t think anyone knows how to spell it) and what ever it lands on that’s the answer lmao. I do that for 99% of the test
I did actually try to solve on my own, I got 5 at first cause I forgot to make the minus into plus after changing the side of some expression. Took me some while to figure out where I got it wrong
Actually he had to talk a lot and stumbled a few times while writing it on board that's why it took him soooo long. It solved it on notebook and took me some 14-15 mins (actually a lot less I think )... But I really leaned some thing new .. the Pascal pyramid
Three years late, but there is a very simple and elegant way. We are given a+b+c = 1 Also a^2+b^2+c^2=2 =>(a+b+c)^2-2(ab+bc+ca)=2 but a+b+c=1 Therefore solving for ab+bc+ca, we have ab+bc+ca = -1/2. Now we can say that a,b,c are roots of the equation x^3-x^2-x/2+q=0 for some constant q. Using vietas relation we obtain that A_n-A_(n-1)-(1/2)(A_(n-2))+q(A_(n-3)) = 0 where A_n=a^n+b^n+c^n. Putting n=3 in the equation and solving we get q =-1/6. Now to get A_5 we need A_4. GeneratingA_4 using the recurrence relation gives us A_4=25/6. Generating A_5 the same way we get A_5=a^5+b^5+c^5=6
@@savaesmek no, you cannot. Even though in this case it yields the right solution, this is merely a coincidence. Are you familiar with the distributive property of multiplication? It states that a(b+c) = ab+ac. For example, 3(5+2) = 3(7) = 21 but also 3(5+2)= 3(5)+3(2)= 15+6=21. So, if you multiply the relations, you get (a^2+b^2+c^2)(a^3+b^3+c^3) = 6. From the distribute law,we have a^2(a^3+b^3+c^3)+b^2(a^3+b^3+c^3) + c^2(a^3+b^3+c^3) = 6 Using the distributive law on each of the terms again, you get a rather messy and long expression that is a ^ 5 + a ^ 3 * b ^ 2 + a ^ 3 * c ^ 2 + a ^ 2 * b ^ 3 + a ^ 2 * c ^ 3 + b ^ 5 + b ^ 3 * c ^ 2 + b ^ 2 * c ^ 3 + c ^ 5 = 6 As you can see, we do have the a^5, b^5 and c^5 terms. However there are a lot of other terms which we dont want. Ofcourse this can be solved this way with algebra and clever use of the right identities, however this is lengthy, and messy making you prone to maling mistakes.
@@cblpu5575 yeah you're right im 14 and i just had an exercise at school where i had to add up relations so i thought you could do the same but i completely forgot what monstrosity of an answer i would get by multiplying
@@cblpu5575 also since you were kind enough to respond could you please tell me what the branch of math with functions and graphs is called? i would like to learn about it before i get to that lesson
at 19:00 its beggar method or multinomial theorem . answer is i guess n+r-1Cr-1 where n are the total things which are to be distributed among r people. please correct if im wrong. My teachers at school taught me this a year ago
@@General12th well..the fact is that this is actually easy. It's not me bragging around. Yeah okay that's long but it's literally a standard exercise with no particular idea behind (still big up to blackpenredpen for doing that video).
Hi! I found an answer to your problem without heavy calculus. Let's consider the polynom P(X)=(X-a)(X-b)(X-c) By developping P, we obtain that P(X)= X^3 - (a+b+c)X² + (ab+ac+bc) X - abc We have : (a+b+c)²=(a²+b²+c²)+2(ab+ac+bc).So, ab+bc+ac=-0.5 Let d= abc . Therfore, P(X)= X^3 - X² -0.5 X -d (*) As a,b and c are the roots of P, then : P(a) = P(b)= P(c) =0 Then , using (*) : 0=P(a)+P(b)+P(c)=(a^3 - a² -0.5 a -d)+(b^3 - b² -0.5 b -d)+(c^3 - c² -0.5 c -d) =(a^3+b^3+c^3)-(a²+b²+c²)-0.5(a+b+c) - 3d = 3 - 2 -0.5*1 -3 d Then d=1/6 Now, let's calculate a*P(a)+b*P(b)+c*P(c) We have then : 0= (a^4+b^4+c^4)-(a^3+b^3+c^3)-0.5(a²+b²+c²) - d (a+b+c) So, a^4+b^4+c^4 = 3+0.5*2+(1/6)*1= 4+1/6= 25/6 Finally, let's calculate a²*P(a)+b²*P(b)+c²*P(c) We have then : 0= (a^5+b^5+c^5)-(a^4+b^4+c^4)-0.5(a^3+b^3+c^3) - d (a²+b²+c²) So, a^5+b^5+c^5 = 25/6 + 0.5*3 +(1/6)*2 = 6 ;) Using this recursive method, we can calculate a^6+b^6+c^6 then a^7+b^7+c^7, etc I hope I did any calculus mistake :)
@@boong0006 Yes. You can apply this method to any degree. In fact, if the problem was to find (a_1)^(n+1)+ (a_2)^(n+1)+ .....+(a_n)^(n+1) such that you know the values of the sums s_n := (a_1)^k+(a_2)^k+...+(a_n)^k for 1 you compute p_{n-1} s_2 + p_{n-1} s_1 + 2 p_{n-2} = 0 >> you compute p_{n-2} s_3 + p_{n-1} s_2 + 2 p_{n-2} s_3 + 3 p_{n-3} =0 >> you compute p_{n-3} . . . s_n + p_{n-1} s_{n-1} + 2 p_{n-2} s_{n-2} + .... + (n-1) p_1 s_1 + n p_n = 0 >> you compute p_n Now, you want to find s_{n+1} : =(a_1)^(n+1) + ...+ (a_n)^(n+1) Let's calculate a_1 * P(a_1)+...+a_n*P(a_n) we have then 0 = s_{n+1} + p_{n-1} s_n + p_{n-2} s_{n-1} + ...+ p_1 s_2 + p_0 s1 Thus, we deduce the value of s_{n+1} To compute s_{n+k} (where k>1) recursively we calculate (a_1)^k * P(a_1) +...+ (a_n)^k * P(a_n). I hope that was clear :) It's a pity that UA-cam doesn't allow mathematical writing.
2:00 You can solve the question by taking all the products of the original formulas that yield a^5, b^5 and c^5 as summands (those are 5 products). Interestingly, you get 5 different types of summands as well (a^5, a^4 * b, a^3 * b^2, a^3 * b * c, a^2 * b^2 * c). Thus, you get a set of 5 linear equations and 5 variables (the number of times the types of summands appear). Solving this so you only get the first type of summand also gives you the solution 6.
Some people were wondering how you could find the value of a⁴+b⁴+c⁴ without tools such as Wolfram Alpha. A few people were actually pretty snippy about it, but it can be done, and if you have the right approach you can do it without too much trouble. Here’s what I did: I started by expanding (a²+b²+c²)² and substituting a²+b²+c² = 2. Some minor rearrangements left me with what I'll call equation A: a⁴+b⁴+c⁴ = 4-2(a²b²+b²c²+a²c²) Next, I expanded (ab+ac+bc)². From the expansion of (a+b+c)² we know that (ab+ac+bc)² = (-1/2)² = 1/4, so we get that 1/4 = a²b² + b²c² + a²c² + 2abc(a+b+c). Substituting a+b+c=1 and the fact from the video that abc=1/6, we get that a²b²+b²c²+a²c²=-1/12 (which happens to also be the sum of all the natural numbers, but I digress). Swap that result back into Equation A above, and a⁴+b⁴+c⁴=4-2(-1/12)=25/6.
blackpenredpen did it in 40 minutes. How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 in less than 3 minutes? Ready for some fun? Okay, stay with me. Let us assume that a,b and c are zeroes of some cubic polynomial f(x). So we have f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now our task is to simply find those coefficients. a+b+c = 1 2(ab+bc+ca) = (a+b+c)² - a² - b² - c² = 1² - 2 = -1 Hence, ab + bc + ca = -1/2 Also we have the following identity: a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Thankfully, we know other quantities so it gives us abc = 1/6 Hence our polynomial is f(x) = x³ - x² - (1/2)x - 1/6 You still here..?? Good! We are almost there. Check your watches. Since 'a' is a zero of f(x), a³ - a² - (1/2)a - 1/6 = 0 After multiplying both sides of this equation by 'a' and a little bit of rearranging, we get: a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 Voila! I appreciate this video nonetheless.
I got it what you said and I believe only top level of people who mastered algebra can solve it and understood it. At least write the values of a+b+c and other from video to make it more easy to understand. Well done btw but all I hope that you haven't used Google for this question. 🤔 I solved it too but I forgot to assume f(x) = x"3 - x"2-........ That you did. I did (a+b+c) (a^3+ b^3 + c^3) which gives a^4.....+c^4 = 3 ;(. Later I thought to do some assumptions like yours and tried but again I forgot to add trinomial formula of abc ;( and i gave up. Btw I did this much calculation in my mind in 3-4 min so I failed 😅.
I just posted this on my Facebook as If 🍋 + 🦆 + 🌳 = 1, and 🍋^2 + 🦆^2 + 🌳^2 = 2, and 🍋^3 + 🦆^3 + 🌳^3 = 3, what's 🍋^5 + 🦆^5 + 🌳^5? I'm curious what answers I'll get.
Funnily enough, he made one mistake in the beginning around 3:08 mark. But somehow ended up getting the right terms in the next step just because of the ongoing pattern.
The Fundamental Theorem of Symmetric Polynomials seems to make this very easy and also pretty straight forward. I have not double checked it yet, but my rough solution seems ok.
Well after you find a ab+bc+ac and abc you can pretty much solve for nth power constructing third degree polynomial with a,b,c as roots (x^3-x^2-1/2x-1/6 in this case) and using Sn-(Sn-1) - 1/2*(Sn-2) - (Sn-3)/6 = 0, where Sn= a^n + b^n + c^n.
Agreed. In fact, it's probably slightly easier still to not actually find abc but instead leave the constant term in the cubic as an unknown constant D, then run the recurrence once to find a^3+b^3+c^3 in terms of D (and hence determine D).
Actually, there is no need to consider the trinomial expansion. In order to achieve the result of abc=1/6, we can just consider the simple equation of (a+b+c)(a^2+b^2+c^2)=2, with the help of the previous result of ab+bc+ca=-1/2 and the given conditions a+b+c=1 and a^3+b^3+c^3=3.
Glad to see someone else used the same method! In case anyone else was curious, here's a little more detail: Given a + b + c = 1 and a^2 + b^2 + c^2 = 2, we can write: 2 = (a + b + c)(a^2 + b^2 + c^2) = a^3 + a(b^2) + a(c^2) + (a^2)b + b^3 + b(c^2) + (a^2)c + (b^2)c + c^3 = a^3 + b^3 + c^3 + ab(a + b) + bc(b + c) + ac(a + c) = a^3 + b^3 + c^3 + ab(1 - c) + bc(1 - a) + ac(1 - b) = a^3 + b^3 + c^3 + ab + bc + ac - 3abc Given ab + bc + ac = -(1/2) and a^3 + b^3 + c^3 = 3, we have: 2 = 3 - (1/2) - 3abc Which yields abc = 1/6
Much easier approach: First get out: a + b + c = 1 ab + bc + ac = -½ abc = ⅙ then define the sequence x_n=a^n+b^n+c^n x_n can be written as a recursive equations x_(n+3)-(1)x_(n+2)+(-1/2)x(n+1)-1/6 x_n =0 since the substitution gives x_n=z^n gives us the cubic with roots a,b,c. We know x_1, x_2, x_3 and then keep plugging in to get x_4 and x_5
Label those 3 equations: 1, 2, and 3. Now rather than trying to tease the sum of 5th powers out of the other power-sums algebraically, I decided to go for the actual solution(s). [Due to symmetry of the givens, any permutation of a, b, and c, is also a solution]. So it becomes pretty clear early on, that a, b, and c can't all be real. So I tried a couple ideas, and this one worked: Let a be real, and (4) b = c* = x + iy, which are complex conjugates of each other. Then Eq.1 requires a + b + c = a + 2x = 1 (5) x = ½(1-a) Eq.2 can then be boiled down to (6) y² = ¾a² - ½a - ¾ and switching which of the 2 y-values to use, merely swaps b and c, because it conjugates them both. Eq.3 will now become a³ + 2x³ - 6xy² = 3, or, using (5) and (6) to eliminate x and y, it can be written (7) a³ - a² - ½a - ⅙ = 0 which has one real root, which by Newton's Method, I get (8) a = 1.43084956624279... From there, we use (5) and (6) to obtain (9) x = -0.215424783121395...; y = 0.26471399125939... which gives b and c, using (4) (10) b = x + iy; c = x - iy So then we can slug out the required sum of 5th powers: a⁵ + b⁵ + c⁵ = a⁵ + b⁵ + (b⁵)* = a⁵ + 2Re(b⁵) = a⁵ + 2x⁵ - 20x³y² + 10xy⁴ = [to be slugged out later] OTOH, I'd really like to see the algebraic method; it's bound to be more elegant, and should also give an explicit, rather than an approximate, answer. OK, watched it, and it was amazing!! I will still come back to verify your answer with the mess that I got ;-) Meanwhile, bravo! And thanks. WAIT!! I think I just got an insight. That cubic in a up there (Eq.7), should give all 3 roots; that is, its 2 complex-conj roots should give b and c. If that's so, then, with coefficients (1, -1, -½, -⅙), we should have a + b + c = 1 ab + bc + ac = -½ abc = ⅙ all of which were verified in your presentation (the 1st one was given, of course). But it's still a lot of work to get from there to the sum of 5th powers... PS: There are a few extremely annoying interruptions by advertising, which is a concentration-killer, and needs to be eliminated somehow, anyhow! I don't know how much control you have over that, if any. Fred
@@blackpenredpen I don't blame you. It's a bad feature of YT; I blame them. I just wondered whether video uploaders have any control over the way YT advertises on their videos. Looks like maybe they don't. ;-( But I don't want that to take anything away from the marvelous presentation you've put up here! Or from posting other long videos, when that's called for. Fred
Well, well, well, at 38:00 we also know from numberphile that -1/12 = 1+2+3+4+5+..., so it doesn't cancel out against the 1/12 toward the end. Therefore the answer is negative infinity, isn't it? :-)
The first equation describes a plane intersecting the coordinate axes at (1,0,0), (0,1,0), and (0,0,1). The second equation describes a sphere of radius sqrt(2) centered at the origin. Their locus of intersection is a circle, centered at (1/3, 1/3, 1/3), with radius sqrt(5/3), and lying in the plane of the first equation. A parametric equation describing this circle can then be tested for when the sum of the cubes of the elements is 3. This yields the solutions (the number of solutions will be some multiple of three).
@@typo691 Its obvious... The left side of the equation has squared terms and the right side is negative. This is only possible if the left side had terms which had the square root of -1.
This reminds me of when I was proving the Marion Walter Theorem via Coordinate Geometry and I had to solve an 8 variable equation and thus produce and inequality afterwards. Awesome content dude!
It was an easy prob! You just made it more complex! Just multiply two equations and solve: (a^2+b^2+c^2)(a^3+b^3+c^3)=6(given) a^5+a^2.b^3+a^2.c^3+b^2.a^3+b^5+b^2.c^3+c^2.a^3+c^2.b^3+c^5=6 ! -> a^5+b^5+c^5+a^2(b^3+c^3)+b^2(a^3+c^3)+c^2(b^3+a^3)=6 ! We can put the value of b^3+c^3 as 3-a^3...From given equation; a^3+b^3+c^3=3 Similarly put values as a^3+c^3=3-b^3 and b^3+a^3=3-c^3 After putting values everything gets simplified and we get our answer 6! :)
Couldnt u just do Σa=1 Σa²=2 Σa³=3 Then use the formulae Σα²=(Σα)²-2Σαβ and Σα³=(Σα)³-3(Σα)(Σαβ)+3αβγ To get Σab and abc and using that and the value of Σa, form a cubic equation with roots a, b and c. Then put that equation in terms of Sn using Roots of Polynomials laws and then solve for S4 and then S5 which would be your answer bcuz u already have S1, S2 and S3. (Sn=a^n + b^n + c^n)
Around minute 32 when you look at the board and said something's wrong shit. That legit is me whenever I get a negative value for time after a page of calculations and now have to go to the beginning look where I lost the sign
There's a theorem about mistakes in long calculations: "The length of time it takes to find your mistake is inversely proportional to the number of possible answers." And the corollary to that, is: "Finding the sign of a result takes the longest possible time." Fred
Usually when u have this kind of questions is very useful using polynomials. Define P(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc. We know that (a+b+c)=1 Doing the same as you did in the video you get that: ab+bc+ca =-1/2. So far P(x)=x^3-x^2-1/2x-abc. We must find abc now. Then, knowing that a,b,c are root of P(x), we have that: P(a)+P(b)+P(c)=0 substituting the values we know we get that abc=1/6. So our polynomial is P(x)=x^3-x^2-1/2x-1/6. Now consider xP(x), with the same trick as before, we get that a^4+b^4+c^4=25/6 And considering x^2P(x) we get that a^5+b^5+c^6=6. Probably you already knew this but i thinks is another approach that is worth sharing ^^
man it was much easier . assume an eqn x³+px²+qx+r=0 with roots a,b,c and apply newtons formul . like put a,b,c in eqn and add them get a relation between p,q,r then multiply eqn by x and repeat then multiply again by x and repeat and get ans easily
A maths professor would have just started with a complicated expression built of (copies of) a+b+c, a²+b²+c² and a³+b³+c³ and miraculously have it condense into a⁵+b⁵+c⁵. We're lucky to see the ways you can actually come up with such solutions.
14:24 => a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac). You could have gotten "abc" value from here. You complicated it unnecessarily. No need for trinomial expansion.
Actually an easy way to do this is to make a third degree polynomial dx^3+ex^2+fx+g=0 having its roots as a, b, c and then using the properties of sum of roots and product of roots we can find a^4+b^4+c^4 this way this question can be done in less time.
This is also a complex equation, we can make a cubic equation whose roots will be a,b and c then we can solve for a, b and c, b and c will be complex numbers and b will be the conjugate of c and a will be the fifth root of 6, then can substitute the values of a,b and c^5 + b^5 will become zero by one of the properties of conjugate of complex numbers the we will get that a^5 = 6.
@@uddeshyaraj2848, Yea, question. On the simple phone, letter division, I choose the figure part (here under knob "?123", back to letters under "ABC"). In this division, (incl.@#£€_&…) press the cipher buttons somewhat longer, like under letter e: èêéëėēę. Under 1 appearing here ¹,½⅓¼⅛, under 2 appearing just ²,⅔, &c. Unfortunately under 6,8,9,0 nothing else turns up here. Well, an ⁿ under 0. So aⁿb²ⁿ possible, but a^(10) · b^6 only like this. The degree ring would look like a¹°. Helps?
I raised (a+b+c)^5 and got interested about the relationship between dimensions (I just call it dimensions now, distinct variables within the parantheses, like this) and the power it is raised to the amount of each term in the expanded result, but I am too lazy to research it, so I just watch the video. Watched part of the video: I actually got 1. abc = 1/6 2. ab + bc + ca = -1/2 but didn't know what to do with it lol. And I got 2. from (a+b+c)(a^2+b^2+c^2) and (a+b+c)^3 instead.
According to wolfram alpha they are 1.43085, -0.215425 + 0.264713 i, and -0.215425 - 0.264713 i. So just raise those in your head to the 5th power, add them together, round off, and you will get 6 ;)
I think that it's literally impossible to solve for an individual letter. Whatever you assign to A, that could be the value for C or B instead. This is because A, B, and C are all to the same power in each line of the equation. So maybe A is a positive number and B is a negative number, maybe vice versa. There's no unique solution.
@@Susanmugen You can solve for the three values but not their order. So you have 6 possible solutions corresponding to the 6 permutations of the three numbers.
3:09 On the seventh term it should be a^3*c^2 Even though on the rest of the process you use the correct term, I just want to metion it because when you wrote a^2*c^2(a+c) on 5:08 I was getting confuse because I wasn't doing the excersie just looking, I was just looking at the terms you have.
@@ΜηδὲνἌγαν-β7φ It's because the solutions are not real, there are no possible real number values of a,b and c. But that doesn't mean there are absolutely no solutions, a, b and c are complex numbers and actually you can easily find them as they are the roots of a cubic equation
Why didnt you just solve it the usual way? By solving the first equation for A, then substitute (1-b-c) for all A's in the second and third equation. Then solve for B, and subsitute the solution for B into the third equation. Then you know, what value C is, and you just similarly have to solve for A and B, which is super easy now. I think, this would have been a much faster way to solve this riddle.
This is not actually faster. I'm doing it now and I'm maybe one third of the way through. There is still a ton of algebra. This solution may make more sense to certain people, though.
You'd still have to square and cube a trinomial, and then solve a system with two equations involving squares and mixed terms. I don't know if that's any faster...
Problems of this sort are easily solved using Gröbner Basis and multivariate polynomial long division. Wolfram Alpha, or Mathematica, or some other computer algebra system will do these calculations. The value of a^5+b^5+c^5 is the remainder after dividing it by the given polynomials, which must be zero. The division algorithm starts with a dividend polynomial, and a set of basis polynomials. It attempts to rewrite the dividend as a linear combination of the basis polynomials (the scalars are polynomials), plus a minimal remainder. Since each basis polynomial is known to be zero, the linear combinations are also known zero. The remainder is congruent to the dividend modulo the basis polynomials. If the set of basis polynomials is the output of the Gröbner Basis algorithm, then the division algorithm is certain to terminate, and the remainder is certain to be minimal. The Gröbner Basis algorithm repeatedly replaces one of the basis polynomials with a linear combination of two or more basis polynomials, until the basis meets the definition of a Gröbner basis. Below, I exhibit the linear combinations which converted the original basis to a Gröbner basis. GroebnerBasis[{ a+b+c-1, a^2+b^2+c^2-2, a^3+b^3+c^3-3}, {a,b,c}] Output: 6c^3 - 6c^2 - 3c - 1 2b^2 + 2bc - 2b + 2c^2 - 2c - 1 a + b + c - 1 Each output polynomial is a linear combination (using polynomials as scalars) of the input polynomials. (a + b + c - 1) = (1)(a + b + c - 1) + (0)(a^2 + b^2 + c^2 - 2) + (0)(a^3+b^3+c^3-3) (2b^2 + 2bc - 2b + 2c^2 - 2c - 1) = (-a + b + c - 1)(a + b + c - 1) + (1)(a^2 + b^2 + c^2 - 2) + (0)(a^3+b^3+c^3-3) (6c^3 - 6c^2 - 3c - 1) = -(a^2 + a (-2 b + c - 1) + b^2 + b (c - 1) - 2 c^2 + 2 c + 1)(a + b + c - 1) - (a + b - 2 c + 2)(a^2+b^2+c^2-2) +(2)(a^3+b^3+c^3-3) To find the value of a^5+b^5+c^5, we rewrite it as a linear combination of the basis polynomials, plus remainder. PolynomialReduce[ a^5+b^5+c^5, {6c^3 - 6c^2 - 3c - 1, 2b^2 + 2bc - 2b + 2c^2 - 2c - 1, a + b + c - 1},{a,b,c}] The output is {{a^2/6 + a/6 + b^2/6 + b/6 + c^2/6 + c/6 + 3/4, a^3/2 - (a^2 c)/2 + a^2/2 - (a c)/2 + (3 a)/4 + b^3/2 - (b^2 c)/2 + b^2/2 - (b c)/2 + (3 b)/4 - (3 c)/2 + 13/6, a^4 - a^3 b - a^3 c + a^3 + a^2 b c - a^2 b - a^2 c + (3 a^2)/2 + a b c - (3 a b)/2 - (3 a c)/2 + (13 a)/6 + (3 b c)/2 - (13 b)/6 - (13 c)/6 + 37/12}, 6} The remainder is 6, as expected. Here is a harder problem, maybe too hard. a + 2b^2 + 3c^3 - 4 = 0 b + 5c^2 + 6a^3 - 7 = 0 c + 8a^2 + 9b^3 -10 = 0 What is the value of a^5 + b^6 + c^7? The answer is the remainder from this division. Unfortunately, Wolfram Alpha chokes on it. PolynomialReduce[ a^5 + b^6 + c^7, GroebnerBasis[ { a + 2b^2 + 3c^3 - 4, b + 5c^2 + 6a^3 - 7, c + 8a^2 + 9b^3 -10}, {a,b,c}], {a,b,c}] The first basis polynomial is 6778308875544 c^27 - 81339706506528 c^24 + 432249688623168 c^21 - 553236458292 c^20 - 1487694677760 c^19 - 1325748264312132 c^18 - 4969645091424 c^17 + 12689140229496 c^16 + 2581342018329456 c^15 + 41556972361824 c^14 - 40066816403646 c^13 - 3316206065297760 c^12 - 110981103029964 c^11 + 62177150238264 c^10 + 2831513992215174 c^9 + 138562705478248 c^8 - 49335765247644 c^7 - 1566288994307279 c^6 - 82771186845288 c^5 + 18441771770903 c^4 + 514283558881536 c^3 + 19636106232323 c^2 - 2175526923685 c - 77174769571225 The others are larger.
The proof seems to be long winding, can be boring to loose patience. My beginnings in Algebra was through my late beloved father and he describes all long proofs as “ANY THING LONG IS WRONG”. But the effort and to keep the old algebra alive by guys like you is much appreciated.. May I offer this solution? Obs 1: (A + B +C)^2 - (A^2 + B^2 + C^2) = 2(AB + BC +CA). Hence AB +BC +CA = (1/2)[ (A + B +C)^2 - (A^2 + B^2 + C^2)] = -(1/2). Obs 2: (A + B + C)(A^2 + B^2 + C^2) = (A^3 + B^3 + C^3) +(A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) = 1.2 = 2. Hence: (A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) = 2 - 3 = -1 Obs 3: (A + B +C)^3 = (A^3 + B^3 + C^3) +3(A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) +6.A.B.C = 3 + 3(-1) +6.A.B.C = 1. Hence: A.B.C = 1/6. Summarizing the observations: A + B + C = 1, AB +BC +CA = -(1/2), A.B.C = 1/6. Hence the cubic equation where A, B and C are roots is given by X^3 - X^2 - (1/2).X - (1/6) = 0. Now if a sequence of numbers is given by by U(N) = A^N + B^N + C^N, it satisfies the relationship U(N + 3) - U(N + 2) - (1/2).U(N + 1) -(1/6)U(N) = 0 is given Clearly U(0) = 3 [Since A^0 + B^0 +C^0 =3], U(1) = 1, U(2) = 2, U(3) = 3, Hence U(4) = U(3) +(1/2)U(2) +(1/6)U(1) = 3 + (1/2).2 + (1/6).(1) = 4 1/6 or 25/6. U(5) = U(4) + (1/2).U(3) + (1/6).U(2) = 25/6 + (1/2).3 + (1/6)(2) = 25/6 + 9/6 + (2/6) = 36/6 = 6 Therefore A^5 + B^5 + C^5 = 6
Please spread this school on your wall and your friends because students need to see it. So will you. There are more than 4 000 tasks and theorems for primary and secondary school students. Thank you for your contribution. Respect! ........................................... .Të lutem shpërndaje këtë shkollë në murin tuaj dhe miqve tuaj, sepse nxënësit kanë nevojë ta shohin. Kështu do të kontribosh edhe ti. Janë më tepër se 4 000 detyra e teorema për nxënës të shkollave të nivelit fillor dhe të mesëm. Faleminderit për kontributin tuaj. Respekt! ua-cam.com/channels/ELXh5_dCYX6umKP3qlOL9g.html
Trinomial theorem reminded me this. My maths professor once said that the general solution is usually the worst in a practical sense. To illustrate the point, he started integrating cos x using Weierstrass substitution. (After substitution t=tan (x/2) you get 2(1-t^2)/(1+t^2)^2 . Have a nice day... :) ) Instead of Trinomial theorem, I think its much easier to use some combinatorical thinking: (a+b+c)^3 = a^3 +b^3 +c^3 + (how many three letter words you can get by taking two equal letters and one different ? =3, that different letter in three positions)*(a^2*b +a^2*c +...) + (what about three different letters ? =6, first one goes in one of three positions * second into one of two remaining)*abc. Then you add coefficients. Is the sum equal to (number of terms)^power ? ( in our case, 1+1+1+3*6 + 6=27=3^3, yay :) )
Inorder to do these type of extreme algebra just remember these newtons formulas S1=a+b+c S2=ab+bc+ac S3=abc Pn=a^n+b^n+c^n 1)P1=S1 2)P2=S1P1-2S2 3)P3=S1P2-S2P1+3S3 4)P4=S1P3-S2P2+S3P1 5)P5=S1P4-S2P3+S3P2 Rest is easy.
To calculate abc it would have been much easier to use the previous result (a+b+c)^2=2*(1+ab+bc+ac) So that (a+b+c)^3=(a+b+c)*2*(1+ab+bc+ac) ==> (a+b+c)+3*abc+(a^2+b^2+c^2)-(a^3+b^3+c^3)=1/2 ==> Abc=1/6
In a math class we were told that in th xy plane if we have a parallelogram and its base has sqrt(something) then the height of that parallelogram will have the same sqrt(of the same thing) Ex: If the base is =2sqrt(13) Height must have a sqrt(13) in it Anyone knows why? Or a proof..! Thx😊
Thirsty For Knowledge ik but the teacher(a great one) told us that when we have a sqrt of a number in a height of the parallelogram then it must be the same sqrt in the base or it will be a miscalculation in our side and we have to check it And after so many times his statement is not false a single time but he doesn't tell us why and want us to search for it
a^2*b^2+a^2*c^2+b^2*c^2 is negative, and so one has to be complex. But since the sum and product are real, another one has to be the conjugate of that one.
the way you arranged the problem in 4:12 reminded me of symmetric matrix diagonalization question. Preety sure its just a coincedence or smth. Great video btw.
EXTREME CHALLENGE:
Find a, b, and c individually!
:/ hold my bear
Next video hopefully?
They are the solutions of a 3th degree polynomial!
no
@@Davidamp 3th
I'm just gonna say 5 and hope the rest of the test is easier.
Same, I just do. Eny meany myny mo( don’t think anyone knows how to spell it) and what ever it lands on that’s the answer lmao. I do that for 99% of the test
What if it’s the only question on the test 🤣🤣🤣
@@Mukawakadoodoo then you are in trouble
Lmao
I did actually try to solve on my own, I got 5 at first cause I forgot to make the minus into plus after changing the side of some expression. Took me some while to figure out where I got it wrong
hes becoming a real professor: occasionally confused by himself
R Hahahhahahaahahaha right on!
Hes becoming a real professor:
@@mystudy6827 isn he?
He is already a teacher lmao
LORD hopefully he learned to slow down so people can understand . It’s not a race.
*pauses the video to try to solve it*
*notices the 41 minutes length*
*plays the video and throws the notebook away*
😂😂😂
I applaud you! Glad I was not the only one who tried to solve it a little.
i triend to solve it then realease it would take for ever to solve the problem and i am getting nothing afterwards
🤣🤣🤣🤣🤣🤣🤣🤣
Actually he had to talk a lot and stumbled a few times while writing it on board that's why it took him soooo long. It solved it on notebook and took me some 14-15 mins (actually a lot less I think )... But I really leaned some thing new .. the Pascal pyramid
Me:*looks at whiteboard*
Whiteboard: 0:30
Me:*looks outside the window for 0.000001seconds
The whiteboard: 40:04
It's a shame the joke has to use timestamps. Wish it could've used pictures. 😂
True
in class everytime 🤣🤣🤣
“Try this first”
Me seeing that this is a 40 minute video: “no, I don’t think I will”
Teachers example:
4x-3y=6
y=-3x+15
Classwork:
x-3y+3z=-4
2x+3y-z=15
4x-3y-z=19
Exam:
Exam: prove a^n + b^n ≠ c^n (n>2)
@@ilya9481 Hahah that's the last theorem of Fermat's, Incredible brain to solve it
@@ilya9481
I actually started solving that.
(5/3, 44/9, 3) (x,y,z)
@@GodOnVayne . Did you use matrix multiplication to work the equation out?
15:55 * Erases K *... No, It's C to the k power * Writes a new K *
Lol
lmao ye and the way he goes "whew" afterwards like thank goodness I noticed. omg my sides
He fixed it in minute 20 (20:00)
Thats so me
That ruined my ocd
"Try it for yourself"
*opens MatLab*
*gets answer with less then 10 lines of code*
B-b-but that's not the point man :(
Code pls
But TBH it ain't hard, it is just complicated
How did you write the code, could you share yours, ive been trying to make it but I just cant:( send yours pls
Three years late, but there is a very simple and elegant way.
We are given a+b+c = 1
Also a^2+b^2+c^2=2
=>(a+b+c)^2-2(ab+bc+ca)=2 but a+b+c=1
Therefore solving for ab+bc+ca, we have ab+bc+ca = -1/2.
Now we can say that a,b,c are roots of the equation
x^3-x^2-x/2+q=0 for some constant q.
Using vietas relation we obtain that A_n-A_(n-1)-(1/2)(A_(n-2))+q(A_(n-3)) = 0 where A_n=a^n+b^n+c^n.
Putting n=3 in the equation and solving we get q =-1/6. Now to get A_5 we need A_4. GeneratingA_4 using the recurrence relation gives us A_4=25/6. Generating A_5 the same way we get A_5=a^5+b^5+c^5=6
cant you just multiply the relations like
a^2+b^2+c^2 = 2
a^3+b^3+c^3 = 3
--------------------------------- (x)
a^5+b^5+c^5 = 3x2 = 6
@@savaesmek no, you cannot. Even though in this case it yields the right solution, this is merely a coincidence.
Are you familiar with the distributive property of multiplication?
It states that a(b+c) = ab+ac. For example, 3(5+2) = 3(7) = 21 but also 3(5+2)= 3(5)+3(2)= 15+6=21.
So, if you multiply the relations, you get (a^2+b^2+c^2)(a^3+b^3+c^3) = 6. From the distribute law,we have
a^2(a^3+b^3+c^3)+b^2(a^3+b^3+c^3) + c^2(a^3+b^3+c^3) = 6
Using the distributive law on each of the terms again, you get a rather messy and long expression that is
a ^ 5 + a ^ 3 * b ^ 2 + a ^ 3 * c ^ 2 + a ^ 2 * b ^ 3 + a ^ 2 * c ^ 3 + b ^ 5 + b ^ 3 * c ^ 2 + b ^ 2 * c ^ 3 + c ^ 5 = 6
As you can see, we do have the a^5, b^5 and c^5 terms. However there are a lot of other terms which we dont want. Ofcourse this can be solved this way with algebra and clever use of the right identities, however this is lengthy, and messy making you prone to maling mistakes.
@@cblpu5575 yeah you're right im 14 and i just had an exercise at school where i had to add up relations so i thought you could do the same but i completely forgot what monstrosity of an answer i would get by multiplying
@@savaesmek haha, happens all the time
@@cblpu5575 also since you were kind enough to respond could you please tell me what the branch of math with functions and graphs is called? i would like to learn about it before i get to that lesson
Mistake at 3:06 . a^2*c^3 should be a^3*c^2 but the rest is okay.
I was going to comment , but ...
15:50 u wrote k no c in base
blackpenredpen How is this written 2 days ago?
@@omarmachuca6405 That video wasn't public
at 19:00 its beggar method or multinomial theorem . answer is i guess n+r-1Cr-1 where n are the total things which are to be distributed among r people. please correct if im wrong. My teachers at school taught me this a year ago
If I ever become a college professor, I'm going to put this as a common exercise and tell my students it's trivial
"The proof is an exercise for the reader."
But... it is actually trivial
@@jahad6335 If the derivation takes forty minutes, it's not trivial. You're just trying to make yourself look smart. Cut it out.
@@General12th One time a professor told me that proofs start being not trivial when they take more than 3 pages xD
@@General12th well..the fact is that this is actually easy. It's not me bragging around. Yeah okay that's long but it's literally a standard exercise with no particular idea behind (still big up to blackpenredpen for doing that video).
That one kid in math class who does this even when the teacher says you don't have to show your work
Please pass the Mustard hahahaha
Xd
0:14 "We know the fourth power is going to be too easy"
Uh, excuse me?
yea it came in one of my exams and i took 10 minutes solving it 😭( i am in 6)
I love how he is always so excited about math! Such an inspiration
Hi! I found an answer to your problem without heavy calculus.
Let's consider the polynom P(X)=(X-a)(X-b)(X-c)
By developping P, we obtain that P(X)= X^3 - (a+b+c)X² + (ab+ac+bc) X - abc
We have : (a+b+c)²=(a²+b²+c²)+2(ab+ac+bc).So, ab+bc+ac=-0.5
Let d= abc . Therfore, P(X)= X^3 - X² -0.5 X -d (*)
As a,b and c are the roots of P, then : P(a) = P(b)= P(c) =0
Then , using (*) : 0=P(a)+P(b)+P(c)=(a^3 - a² -0.5 a -d)+(b^3 - b² -0.5 b -d)+(c^3 - c² -0.5 c -d)
=(a^3+b^3+c^3)-(a²+b²+c²)-0.5(a+b+c) - 3d
= 3 - 2 -0.5*1 -3 d
Then d=1/6
Now, let's calculate a*P(a)+b*P(b)+c*P(c)
We have then : 0= (a^4+b^4+c^4)-(a^3+b^3+c^3)-0.5(a²+b²+c²) - d (a+b+c)
So, a^4+b^4+c^4 = 3+0.5*2+(1/6)*1= 4+1/6= 25/6
Finally, let's calculate a²*P(a)+b²*P(b)+c²*P(c)
We have then : 0= (a^5+b^5+c^5)-(a^4+b^4+c^4)-0.5(a^3+b^3+c^3) - d (a²+b²+c²)
So, a^5+b^5+c^5 = 25/6 + 0.5*3 +(1/6)*2 = 6 ;)
Using this recursive method, we can calculate a^6+b^6+c^6 then a^7+b^7+c^7, etc
I hope I did any calculus mistake :)
Thank you!
Since P(X) is a trinomial, I wonder if you can apply this method to a quadnomial that involves a,b,c and d?
Thats the best comment ever!!
In the last line you got a mistake. It is not 16/3 but 6. This is an amazing method, because you can find other sum of powers
@@boong0006 Yes. You can apply this method to any degree. In fact, if the problem was to find (a_1)^(n+1)+ (a_2)^(n+1)+ .....+(a_n)^(n+1) such that you know the values of the sums s_n := (a_1)^k+(a_2)^k+...+(a_n)^k for 1 you compute p_{n-1}
s_2 + p_{n-1} s_1 + 2 p_{n-2} = 0 >> you compute p_{n-2}
s_3 + p_{n-1} s_2 + 2 p_{n-2} s_3 + 3 p_{n-3} =0 >> you compute p_{n-3}
.
.
.
s_n + p_{n-1} s_{n-1} + 2 p_{n-2} s_{n-2} + .... + (n-1) p_1 s_1 + n p_n = 0 >> you compute p_n
Now, you want to find s_{n+1} : =(a_1)^(n+1) + ...+ (a_n)^(n+1)
Let's calculate a_1 * P(a_1)+...+a_n*P(a_n)
we have then 0 = s_{n+1} + p_{n-1} s_n + p_{n-2} s_{n-1} + ...+ p_1 s_2 + p_0 s1
Thus, we deduce the value of s_{n+1}
To compute s_{n+k} (where k>1) recursively we calculate (a_1)^k * P(a_1) +...+ (a_n)^k * P(a_n).
I hope that was clear :) It's a pity that UA-cam doesn't allow mathematical writing.
Skipping from 0:00 to 28:00
Is like my teacher after i took a restroom.
@@codeguy21 dafaq
@@codeguy21 wut?
@@codeguy21 TF?
damn bro pass me a restroom too
that means you go to the restroom for 28 minutes.
Congrats on showing the non-edited video. We've all been there, batteling hard algebra. I enjoyed seeing this video. ☺
Mauro Maia thank you!!!!
@@blackpenredpen I found a way to solve this problem which is very easy and short
Anshuman Agrawal Please share
@@dux2508 Simply use Newton's Identities and consider a,b, and c as roots of a cubic
@@MrKrabs-xf2tr Yes, exactly
2:00 You can solve the question by taking all the products of the original formulas that yield a^5, b^5 and c^5 as summands (those are 5 products).
Interestingly, you get 5 different types of summands as well (a^5, a^4 * b, a^3 * b^2, a^3 * b * c, a^2 * b^2 * c).
Thus, you get a set of 5 linear equations and 5 variables (the number of times the types of summands appear). Solving this so you only get the first type of summand also gives you the solution 6.
Some people were wondering how you could find the value of a⁴+b⁴+c⁴ without tools such as Wolfram Alpha. A few people were actually pretty snippy about it, but it can be done, and if you have the right approach you can do it without too much trouble. Here’s what I did:
I started by expanding (a²+b²+c²)² and substituting a²+b²+c² = 2. Some minor rearrangements left me with what I'll call equation A:
a⁴+b⁴+c⁴ = 4-2(a²b²+b²c²+a²c²)
Next, I expanded (ab+ac+bc)². From the expansion of (a+b+c)² we know that (ab+ac+bc)² = (-1/2)² = 1/4, so we get that 1/4 = a²b² + b²c² + a²c² + 2abc(a+b+c). Substituting a+b+c=1 and the fact from the video that abc=1/6, we get that a²b²+b²c²+a²c²=-1/12 (which happens to also be the sum of all the natural numbers, but I digress).
Swap that result back into Equation A above, and a⁴+b⁴+c⁴=4-2(-1/12)=25/6.
How interesting, it's approximately 4
blackpenredpen did it in 40 minutes. How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 in less than 3 minutes? Ready for some fun? Okay, stay with me.
Let us assume that a,b and c are zeroes of some cubic polynomial f(x). So we have f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc
Now our task is to simply find those coefficients.
a+b+c = 1
2(ab+bc+ca) = (a+b+c)² - a² - b² - c² = 1² - 2 = -1
Hence, ab + bc + ca = -1/2
Also we have the following identity:
a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
Thankfully, we know other quantities so it gives us abc = 1/6
Hence our polynomial is f(x) = x³ - x² - (1/2)x - 1/6
You still here..?? Good! We are almost there. Check your watches.
Since 'a' is a zero of f(x), a³ - a² - (1/2)a - 1/6 = 0
After multiplying both sides of this equation by 'a' and a little bit of rearranging, we get:
a⁴ = a³ + (1/2)a² + (1/6)a
We will get similar equations for b and c.
After adding them,
a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c)
= 3 + 2/2 + 1/6 = 25/6
Similarly, a^5 = a⁴ + (1/2)a³ + (1/6)a²
Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6
Voila!
I appreciate this video nonetheless.
Yea I was taught this way in high school and it blew my mind for a solid week because I did an entire excersise without it and died of the monotony
@@peterderias5323 That's interesting :)
I didn't get it😥
@@gekkouga5147 Have you studied polynomials? If not, I would suggest to study that first. If yes, tell me which part is troubling you.
I got it what you said and I believe only top level of people who mastered algebra can solve it and understood it. At least write the values of a+b+c and other from video to make it more easy to understand. Well done btw but all I hope that you haven't used Google for this question. 🤔 I solved it too but I forgot to assume f(x) = x"3 - x"2-........ That you did. I did (a+b+c) (a^3+ b^3 + c^3) which gives a^4.....+c^4 = 3 ;(. Later I thought to do some assumptions like yours and tried but again I forgot to add trinomial formula of abc ;( and i gave up. Btw I did this much calculation in my mind in 3-4 min so I failed 😅.
I just posted this on my Facebook as If 🍋 + 🦆 + 🌳 = 1, and 🍋^2 + 🦆^2 + 🌳^2 = 2, and 🍋^3 + 🦆^3 + 🌳^3 = 3, what's 🍋^5 + 🦆^5 + 🌳^5?
I'm curious what answers I'll get.
Joshua Hillerup hahahhahahahha
Brilliant.
Winner 🙈
For sure they change your symbols by a, b, c, or x, y, z, etc.
Wait you MUST NOT forget to write 98% fails to solve
Proud to admit I watched the video all the way through instead of doing my homework
bprp: "find a, b, and c individually!"
me: *no*
And what sucks is that if you make *one* mistake, everything to follow is all wrong.
Funnily enough, he made one mistake in the beginning around 3:08 mark. But somehow ended up getting the right terms in the next step just because of the ongoing pattern.
That's actually a good comment George :)
The Fundamental Theorem of Symmetric Polynomials seems to make this very easy and also pretty straight forward. I have not double checked it yet, but my rough solution seems ok.
Well after you find a ab+bc+ac and abc you can pretty much solve for nth power constructing third degree polynomial with a,b,c as roots (x^3-x^2-1/2x-1/6 in this case) and using Sn-(Sn-1) - 1/2*(Sn-2) - (Sn-3)/6 = 0, where Sn= a^n + b^n + c^n.
Agreed. In fact, it's probably slightly easier still to not actually find abc but instead leave the constant term in the cubic as an unknown constant D, then run the recurrence once to find a^3+b^3+c^3 in terms of D (and hence determine D).
I too did with the same method .
@Cool Dude I am from 10th grade too so they probably may :)
Actually, there is no need to consider the trinomial expansion. In order to achieve the result of abc=1/6, we can just consider the simple equation of (a+b+c)(a^2+b^2+c^2)=2, with the help of the previous result of ab+bc+ca=-1/2 and the given conditions a+b+c=1 and a^3+b^3+c^3=3.
Glad to see someone else used the same method! In case anyone else was curious, here's a little more detail:
Given a + b + c = 1 and a^2 + b^2 + c^2 = 2, we can write:
2 = (a + b + c)(a^2 + b^2 + c^2)
= a^3 + a(b^2) + a(c^2)
+ (a^2)b + b^3 + b(c^2)
+ (a^2)c + (b^2)c + c^3
= a^3 + b^3 + c^3
+ ab(a + b)
+ bc(b + c)
+ ac(a + c)
= a^3 + b^3 + c^3
+ ab(1 - c)
+ bc(1 - a)
+ ac(1 - b)
= a^3 + b^3 + c^3
+ ab + bc + ac
- 3abc
Given ab + bc + ac = -(1/2) and a^3 + b^3 + c^3 = 3, we have:
2 = 3 - (1/2) - 3abc
Which yields abc = 1/6
Much easier approach:
First get out:
a + b + c = 1
ab + bc + ac = -½
abc = ⅙
then define the sequence x_n=a^n+b^n+c^n
x_n can be written as a recursive equations x_(n+3)-(1)x_(n+2)+(-1/2)x(n+1)-1/6 x_n =0
since the substitution gives x_n=z^n gives us the cubic with roots a,b,c.
We know x_1, x_2, x_3 and then keep plugging in to get x_4 and x_5
1)Subtract eqn 3 and 1.
2) Equate the terms with eqn2.
3)U get 3 solutions for x , those are for x,y,z
4) profit
Label those 3 equations: 1, 2, and 3.
Now rather than trying to tease the sum of 5th powers out of the other power-sums algebraically, I decided to go for the actual solution(s).
[Due to symmetry of the givens, any permutation of a, b, and c, is also a solution].
So it becomes pretty clear early on, that a, b, and c can't all be real.
So I tried a couple ideas, and this one worked:
Let a be real, and
(4) b = c* = x + iy,
which are complex conjugates of each other. Then Eq.1 requires
a + b + c = a + 2x = 1
(5) x = ½(1-a)
Eq.2 can then be boiled down to
(6) y² = ¾a² - ½a - ¾
and switching which of the 2 y-values to use, merely swaps b and c, because it conjugates them both.
Eq.3 will now become
a³ + 2x³ - 6xy² = 3, or, using (5) and (6) to eliminate x and y, it can be written
(7) a³ - a² - ½a - ⅙ = 0
which has one real root, which by Newton's Method, I get
(8) a = 1.43084956624279...
From there, we use (5) and (6) to obtain
(9) x = -0.215424783121395...; y = 0.26471399125939...
which gives b and c, using (4)
(10) b = x + iy; c = x - iy
So then we can slug out the required sum of 5th powers:
a⁵ + b⁵ + c⁵ = a⁵ + b⁵ + (b⁵)* = a⁵ + 2Re(b⁵) = a⁵ + 2x⁵ - 20x³y² + 10xy⁴ = [to be slugged out later]
OTOH, I'd really like to see the algebraic method; it's bound to be more elegant, and should also give an explicit, rather than an approximate, answer.
OK, watched it, and it was amazing!!
I will still come back to verify your answer with the mess that I got ;-)
Meanwhile, bravo! And thanks.
WAIT!! I think I just got an insight. That cubic in a up there (Eq.7), should give all 3 roots; that is, its 2 complex-conj roots should give b and c.
If that's so, then, with coefficients (1, -1, -½, -⅙), we should have
a + b + c = 1
ab + bc + ac = -½
abc = ⅙
all of which were verified in your presentation (the 1st one was given, of course).
But it's still a lot of work to get from there to the sum of 5th powers...
PS: There are a few extremely annoying interruptions by advertising, which is a concentration-killer, and needs to be eliminated somehow, anyhow!
I don't know how much control you have over that, if any.
Fred
Sorry for the ads. This video is longer than usual.
@@blackpenredpen I don't blame you. It's a bad feature of YT; I blame them.
I just wondered whether video uploaders have any control over the way YT advertises on their videos.
Looks like maybe they don't. ;-(
But I don't want that to take anything away from the marvelous presentation you've put up here!
Or from posting other long videos, when that's called for.
Fred
Hi Fred. To avoid the ads, go to the end of the video and then do "replay".
@@M-F-H Thanks! I'll try that.
Fred
Well, well, well, at 38:00 we also know from numberphile that -1/12 = 1+2+3+4+5+..., so it doesn't cancel out against the 1/12 toward the end. Therefore the answer is negative infinity, isn't it? :-)
No
That's wrong
Awesome! Your exercises are getting more and more interesting!
We can get abc more easily by using
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc
Opens video with "lets do an extereme algebra question for fun". This is the youth culture I have been looking for.
Just watched my 3rd back-to-back blackpenredpen video. My brain now needs a vacation on a tropical island.
Andy awww : ))))
What the hell is this.
vtt355
It’s algebra 2 but like the accelerated version.
Craziness
Our Worst dream.
Abdulaziz Alquhaibi عبدالعزيز هذا شنو 😂
A nightmare
*If this was my exam, it would just be worth like 6 marks, nothing else*
Very easy if you know
a^3+b^3+c^3 -3abc=(a+b+c)(a^ 2b^2 c^2.-ab -bc -ca )
The first equation describes a plane intersecting the coordinate axes at (1,0,0), (0,1,0), and (0,0,1). The second equation describes a sphere of radius sqrt(2) centered at the origin. Their locus of intersection is a circle, centered at (1/3, 1/3, 1/3), with radius sqrt(5/3), and lying in the plane of the first equation. A parametric equation describing this circle can then be tested for when the sum of the cubes of the elements is 3. This yields the solutions (the number of solutions will be some multiple of three).
38:00 implies complex solutions! 😄
duggydo pretty much!
What makes you say so? Experience or memes?
Typo ???
@@typo691 Its obvious... The left side of the equation has squared terms and the right side is negative. This is only possible if the left side had terms which had the square root of -1.
@duggydo, From the first minute we thought of a kind of symmetrical couple, triple, in the complex plane.
You should invest into a small microphone that you can attach to your clothe.
na its iconic at this point
It is a grenade. It si not a microphone. Ja ja ja
@@luismariabiaggioni8514 lmao
Ok
NAH! then how he will show off his black pen red pen trick?
This reminds me of when I was proving the Marion Walter Theorem via Coordinate Geometry and I had to solve an 8 variable equation and thus produce and inequality afterwards. Awesome content dude!
It was an easy prob! You just made it more complex! Just multiply two equations and solve:
(a^2+b^2+c^2)(a^3+b^3+c^3)=6(given)
a^5+a^2.b^3+a^2.c^3+b^2.a^3+b^5+b^2.c^3+c^2.a^3+c^2.b^3+c^5=6 !
-> a^5+b^5+c^5+a^2(b^3+c^3)+b^2(a^3+c^3)+c^2(b^3+a^3)=6 !
We can put the value of b^3+c^3 as 3-a^3...From given equation; a^3+b^3+c^3=3
Similarly put values as a^3+c^3=3-b^3 and b^3+a^3=3-c^3
After putting values everything gets simplified and we get our answer 6!
:)
Couldnt u just do
Σa=1
Σa²=2
Σa³=3
Then use the formulae
Σα²=(Σα)²-2Σαβ
and
Σα³=(Σα)³-3(Σα)(Σαβ)+3αβγ
To get Σab and abc and using that and the value of Σa, form a cubic equation with roots a, b and c.
Then put that equation in terms of Sn using Roots of Polynomials laws and then solve for S4 and then S5 which would be your answer bcuz u already have S1, S2 and S3. (Sn=a^n + b^n + c^n)
Around minute 32 when you look at the board and said something's wrong shit. That legit is me whenever I get a negative value for time after a page of calculations and now have to go to the beginning look where I lost the sign
David Nobre don’t worry, time is relative
hahaha
That's when I inconspicuously add/drop a negative somewhere in my work to just make it work out.
@David Nobre,
Algebra is outside time domain,
algebra is beyond time,
algebra is timeless.
Examinations aren't.
Algebra books aren't.
Olympiad sessions aren't.
Even
blackpenredpen videos aren't.
Regrettable?
Well, uhh, rr.
**
There's a theorem about mistakes in long calculations:
"The length of time it takes to find your mistake is inversely proportional to the number of possible answers."
And the corollary to that, is:
"Finding the sign of a result takes the longest possible time."
Fred
...
I only realised I just watched 40 minutes of algebra after the video finished
Math is immersive. Can't believe it was 40 minutes...
: ))))))
Can't believe it was only 40 minutes. Would watch for a couple of hours.
20 minutes if you 2X it
FROM 15:47 TO 15:53...MY ASIAN DUDE SAYS THAT THERE SHOULD BE C INSTEAD OF 'k' BUT THEN RUBS IT TO WRITE 'k' AGAIN......LMAO
just like everybodys teachers xD
20:06 fixed
Lag
31:00 You can easily find abc=1/6 by factorising a^3+b3+c3 =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc = 3
there was no need for the trinomial expansion.
there is an easier formula:
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - (ab + bc + ac))
Idk if you are right but congrats for even knowing what he's talking about.
I love this guy! He makes math more entertaining than a lot of Hollywood movies I've seen lately!
Usually when u have this kind of questions is very useful using polynomials.
Define P(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc.
We know that (a+b+c)=1
Doing the same as you did in the video you get that: ab+bc+ca =-1/2.
So far P(x)=x^3-x^2-1/2x-abc.
We must find abc now.
Then, knowing that a,b,c are root of P(x), we have that:
P(a)+P(b)+P(c)=0 substituting the values we know we get that abc=1/6.
So our polynomial is P(x)=x^3-x^2-1/2x-1/6.
Now consider xP(x), with the same trick as before, we get that a^4+b^4+c^4=25/6
And considering x^2P(x) we get that a^5+b^5+c^6=6.
Probably you already knew this but i thinks is another approach that is worth sharing ^^
How can i save your solution?😊
@@AnuragGuptainspired Take a screenshot
imagine that the camera werent recording
man it was much easier . assume an eqn x³+px²+qx+r=0 with roots a,b,c and apply newtons formul . like put a,b,c in eqn and add them
get a relation between p,q,r
then multiply eqn by x and repeat
then multiply again by x and repeat and get ans easily
A maths professor would have just started with a complicated expression built of (copies of) a+b+c, a²+b²+c² and a³+b³+c³ and miraculously have it condense into a⁵+b⁵+c⁵.
We're lucky to see the ways you can actually come up with such solutions.
14:24 => a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac).
You could have gotten "abc" value from here. You complicated it unnecessarily. No need for trinomial expansion.
That’s what I thought too!
using the same mathod, a^4+b^4+c^4=25/6
Thats almost 4
Actually an easy way to do this is to make a third degree polynomial dx^3+ex^2+fx+g=0 having its roots as a, b, c and then using the properties of sum of roots and product of roots we can find a^4+b^4+c^4 this way this question can be done in less time.
@@ammarbarbhaiwala9908 wie zum fick kommt einer Mit mois Profilbild unter so ein mathe Video? XD
@@fiNitEarth Was für ein Typ? Das ist JC Denton
@@AndDiracisHisProphet there is no god
Patience is key! Love these kind of videos man, keep it up!
14:36 I love the Trinomial Expansion!!!
Cool solution 🤩
This is also a complex equation, we can make a cubic equation whose roots will be a,b and c then we can solve for a, b and c, b and c will be complex numbers and b will be the conjugate of c and a will be the fifth root of 6, then can substitute the values of a,b and c^5 + b^5 will become zero by one of the properties of conjugate of complex numbers the we will get that a^5 = 6.
"Who tf would watch a 40 minute video of a guy doing algebra???"
-Me, 40 minutes ago
Haha same
i thought answer would be -1/12
@X15 CYBER RUSH,
Just try
a¹² + b¹² + c¹²
You will be surprised.
We were too.
Good luck!
@@keescanalfp5143 how did u write 12 in superscript
@@uddeshyaraj2848,
Yea, question. On the simple phone, letter division, I choose the figure part
(here under knob "?123", back to letters under "ABC"). In this division, (incl.@#£€_&…) press the cipher buttons somewhat longer, like under letter e: èêéëėēę.
Under 1 appearing here ¹,½⅓¼⅛,
under 2 appearing just ²,⅔, &c. Unfortunately under 6,8,9,0 nothing else turns up here. Well, an ⁿ under 0.
So aⁿb²ⁿ possible, but a^(10) · b^6 only like this. The degree ring would look like a¹°. Helps?
@@uddeshyaraj2848 what about using the app Unicode pad
@@uddeshyaraj2848 hold the 1 and 2
"Please pause the video and try this first."
Oh that's alright I'm alrighty finished
you can do it by newton sum by assuming a,b,c as roots of a cubic polynomial
Don’t stress easy way:9a^3> and now convert it down
8:16 that's literally me. I do always solve algebra and get a unexpected result.
16:00 he again write k in place of c😅😅😁
I raised (a+b+c)^5 and got interested about the relationship between dimensions (I just call it dimensions now, distinct variables within the parantheses, like this) and the power it is raised to the amount of each term in the expanded result, but I am too lazy to research it, so I just watch the video.
Watched part of the video:
I actually got
1. abc = 1/6
2. ab + bc + ca = -1/2
but didn't know what to do with it lol.
And I got 2. from (a+b+c)(a^2+b^2+c^2) and (a+b+c)^3 instead.
: )))
I didn't want to do that since (a+b+c)^5 is going to be crazy
If it just to find , you can easily do it by creating a cubic whose roots are a , b , and c
I remember watching it about 2 years back. Man, how much have I grown, how much has changed from then
Ok now solve for a,b and c
According to wolfram alpha they are 1.43085, -0.215425 + 0.264713 i, and -0.215425 - 0.264713 i. So just raise those in your head to the 5th power, add them together, round off, and you will get 6 ;)
I thought that's what was being shown, but was dissapointed :( is it hard to simultaniously solve them?
I think that it's literally impossible to solve for an individual letter. Whatever you assign to A, that could be the value for C or B instead. This is because A, B, and C are all to the same power in each line of the equation. So maybe A is a positive number and B is a negative number, maybe vice versa. There's no unique solution.
believe in math, not wolfram alpha!
@@Susanmugen You can solve for the three values but not their order. So you have 6 possible solutions corresponding to the 6 permutations of the three numbers.
3:09 On the seventh term it should be a^3*c^2 Even though on the rest of the process you use the correct term, I just want to metion it because when you wrote a^2*c^2(a+c) on 5:08 I was getting confuse because I wasn't doing the excersie just looking, I was just looking at the terms you have.
Every Algebra person: Yeah I can do Algebra it's super easy I can do it in my head.
BlackpenRedpen: Hold my beer
Hahaha
Man, what a perfect example of what it feels like when the teacher asks you to show your work, when you got your answer based off your intuition
3b1b for visualizing maths.
BpRp for learning maths.
Thats more than enough.
This is so amazing I’m so surprised
I tried this question with newton's identities by making a polynomial with roots a b and c and got the right answer
That's what I was looking for
Yeahhh! That's what I did too! It helped me solve the question in like 3 minutes or something. It even helped me get the values of a, b, and c!
I wanna know how (ab)^2 + (bc)^2 + (ca)^2 is -1/12
Exactly! The solution is fake. Because a sum of positives makes an negative! This means that the supposes at the begining is impossible!
@@ΜηδὲνἌγαν-β7φ It's because the solutions are not real, there are no possible real number values of a,b and c. But that doesn't mean there are absolutely no solutions, a, b and c are complex numbers and actually you can easily find them as they are the roots of a cubic equation
Better method can be by writing recurrence relation
Brilliant working out impromptu on board. Really good!!
-1/12!
Minus 1/12 factorial? Oh boy
@@gregoritsen ikr
approximately 1.055546564813466302313701484734262209810379114858428728414...
Ikr
@@maxsch.6555 that is more precise than pi
Why didnt you just solve it the usual way?
By solving the first equation for A, then substitute (1-b-c) for all A's in the second and third equation.
Then solve for B, and subsitute the solution for B into the third equation.
Then you know, what value C is, and you just similarly have to solve for A and B, which is super easy now.
I think, this would have been a much faster way to solve this riddle.
This is not actually faster. I'm doing it now and I'm maybe one third of the way through. There is still a ton of algebra. This solution may make more sense to certain people, though.
No it'll be messy
You'd still have to square and cube a trinomial, and then solve a system with two equations involving squares and mixed terms. I don't know if that's any faster...
I started doing exactly that while watching video. I continued today and got
6c³-6c²-3c-1=0 It was not easy to get that. I will not continue.
@@jkn6644 solve it using Cardan's method (or put it on Wolfram alpha)
Problems of this sort are easily solved using Gröbner Basis and multivariate polynomial long division. Wolfram Alpha, or Mathematica, or some other computer algebra system will do these calculations.
The value of a^5+b^5+c^5 is the remainder after dividing it by the given polynomials, which must be zero.
The division algorithm starts with a dividend polynomial, and a set of basis polynomials. It attempts to rewrite the dividend as a linear combination of the basis polynomials (the scalars are polynomials), plus a minimal remainder.
Since each basis polynomial is known to be zero, the linear combinations are also known zero. The remainder is congruent to the dividend modulo the basis polynomials.
If the set of basis polynomials is the output of the Gröbner Basis algorithm, then the division algorithm is certain to terminate, and the remainder is certain to be minimal.
The Gröbner Basis algorithm repeatedly replaces one of the basis polynomials with a linear combination of two or more basis polynomials, until the basis meets the definition of a Gröbner basis. Below, I exhibit the linear combinations which converted the original basis to a Gröbner basis.
GroebnerBasis[{
a+b+c-1,
a^2+b^2+c^2-2,
a^3+b^3+c^3-3},
{a,b,c}]
Output:
6c^3 - 6c^2 - 3c - 1
2b^2 + 2bc - 2b + 2c^2 - 2c - 1
a + b + c - 1
Each output polynomial is a linear combination (using polynomials as scalars) of the input polynomials.
(a + b + c - 1) =
(1)(a + b + c - 1)
+ (0)(a^2 + b^2 + c^2 - 2)
+ (0)(a^3+b^3+c^3-3)
(2b^2 + 2bc - 2b + 2c^2 - 2c - 1) =
(-a + b + c - 1)(a + b + c - 1)
+ (1)(a^2 + b^2 + c^2 - 2)
+ (0)(a^3+b^3+c^3-3)
(6c^3 - 6c^2 - 3c - 1) =
-(a^2 + a (-2 b + c - 1) + b^2 + b (c - 1) - 2 c^2 + 2 c + 1)(a + b + c - 1)
- (a + b - 2 c + 2)(a^2+b^2+c^2-2)
+(2)(a^3+b^3+c^3-3)
To find the value of a^5+b^5+c^5, we rewrite it as a linear combination of the basis polynomials, plus remainder.
PolynomialReduce[ a^5+b^5+c^5, {6c^3 - 6c^2 - 3c - 1, 2b^2 + 2bc - 2b + 2c^2 - 2c - 1, a + b + c - 1},{a,b,c}]
The output is
{{a^2/6 + a/6 + b^2/6 + b/6 + c^2/6 + c/6 + 3/4, a^3/2 - (a^2 c)/2 + a^2/2 - (a c)/2 + (3 a)/4 + b^3/2 - (b^2 c)/2 + b^2/2 - (b c)/2 + (3 b)/4 - (3 c)/2 + 13/6, a^4 - a^3 b - a^3 c + a^3 + a^2 b c - a^2 b - a^2 c + (3 a^2)/2 + a b c - (3 a b)/2 - (3 a c)/2 + (13 a)/6 + (3 b c)/2 - (13 b)/6 - (13 c)/6 + 37/12}, 6}
The remainder is 6, as expected.
Here is a harder problem, maybe too hard.
a + 2b^2 + 3c^3 - 4 = 0
b + 5c^2 + 6a^3 - 7 = 0
c + 8a^2 + 9b^3 -10 = 0
What is the value of a^5 + b^6 + c^7?
The answer is the remainder from this division. Unfortunately, Wolfram Alpha chokes on it.
PolynomialReduce[
a^5 + b^6 + c^7,
GroebnerBasis[ {
a + 2b^2 + 3c^3 - 4,
b + 5c^2 + 6a^3 - 7,
c + 8a^2 + 9b^3 -10},
{a,b,c}],
{a,b,c}]
The first basis polynomial is
6778308875544 c^27 - 81339706506528 c^24 + 432249688623168 c^21 - 553236458292 c^20 - 1487694677760 c^19 - 1325748264312132 c^18 - 4969645091424 c^17 + 12689140229496 c^16 + 2581342018329456 c^15 + 41556972361824 c^14 - 40066816403646 c^13 - 3316206065297760 c^12 - 110981103029964 c^11 + 62177150238264 c^10 + 2831513992215174 c^9 + 138562705478248 c^8 - 49335765247644 c^7 - 1566288994307279 c^6 - 82771186845288 c^5 + 18441771770903 c^4 + 514283558881536 c^3 + 19636106232323 c^2 - 2175526923685 c - 77174769571225
The others are larger.
The proof seems to be long winding, can be boring to loose patience. My beginnings in Algebra was through my late beloved father and he describes all long proofs as “ANY THING LONG IS WRONG”. But the effort and to keep the old algebra alive by guys like you is much appreciated.. May I offer this solution?
Obs 1: (A + B +C)^2 - (A^2 + B^2 + C^2) = 2(AB + BC +CA).
Hence AB +BC +CA = (1/2)[ (A + B +C)^2 - (A^2 + B^2 + C^2)] = -(1/2).
Obs 2: (A + B + C)(A^2 + B^2 + C^2) = (A^3 + B^3 + C^3) +(A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) = 1.2 = 2.
Hence: (A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) = 2 - 3 = -1
Obs 3: (A + B +C)^3 = (A^3 + B^3 + C^3) +3(A^2.B + B^2.C + C^2.A + A^2.C + C^2.B + B^2.A) +6.A.B.C = 3 + 3(-1) +6.A.B.C = 1.
Hence: A.B.C = 1/6.
Summarizing the observations:
A + B + C = 1, AB +BC +CA = -(1/2), A.B.C = 1/6.
Hence the cubic equation where A, B and C are roots is given by
X^3 - X^2 - (1/2).X - (1/6) = 0.
Now if a sequence of numbers is given by by U(N) = A^N + B^N + C^N, it satisfies the relationship U(N + 3) - U(N + 2) - (1/2).U(N + 1) -(1/6)U(N) = 0 is given
Clearly U(0) = 3 [Since A^0 + B^0 +C^0 =3],
U(1) = 1, U(2) = 2, U(3) = 3,
Hence U(4) = U(3) +(1/2)U(2) +(1/6)U(1) = 3 + (1/2).2 + (1/6).(1) = 4 1/6 or 25/6.
U(5) = U(4) + (1/2).U(3) + (1/6).U(2) = 25/6 + (1/2).3 + (1/6)(2) = 25/6 + 9/6 + (2/6) = 36/6 = 6
Therefore A^5 + B^5 + C^5 = 6
Very Nice Method Of Teaching .Stay Blessed .
I was like "Bruh stop" when he said "Because 3x2 is equal to 6"
and then silence, and he said "Because 1+2+3=6" and I suddenly laughed.
*Dude why*
Lolllll
@@blackpenredpen Fun fact İ was trying to solve this in my dream. Thank you for making it a nightmare.
When you realise that actually isn't a joke...
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5:11 a^2 c^3 plus a^2c^3 doesn't factorise to a^2c^2 (a+c) That part I know is quite incorrect!
I’ve never heard him swear before. That was hilarious
@Memes shorts
I = You
Wait when did he swear
@@78anurag
When he freaked out
@@fanamatakecick97 Timestamp?
@@78anurag
Before he did (ab + bc + ac)^2
In the first observation, 3rd line, instead of (a^2)(c^3) isnt it supposed to be (a^3)(c^2)?
You can use a³+b³+c³=(a+b+c)³-3(a+b)(b+c)(c+a) to get value of (abc).
Trinomial theorem reminded me this. My maths professor once said that the general solution is usually the worst in a practical sense. To illustrate the point, he started integrating cos x using Weierstrass substitution. (After substitution t=tan (x/2) you get 2(1-t^2)/(1+t^2)^2 . Have a nice day... :) )
Instead of Trinomial theorem, I think its much easier to use some combinatorical thinking: (a+b+c)^3 = a^3 +b^3 +c^3 + (how many three letter words you can get by taking two equal letters and one different ? =3, that different letter in three positions)*(a^2*b +a^2*c +...) + (what about three different letters ? =6, first one goes in one of three positions * second into one of two remaining)*abc. Then you add coefficients. Is the sum equal to (number of terms)^power ? ( in our case, 1+1+1+3*6 + 6=27=3^3, yay :) )
You know its a fun question when even bprp gets confused
How many times you wanna be confused with your own answer ?
him: yes. eh? wait..oh yeah yes. it's yes.
Inorder to do these type of extreme algebra
just remember these newtons formulas
S1=a+b+c
S2=ab+bc+ac
S3=abc
Pn=a^n+b^n+c^n
1)P1=S1
2)P2=S1P1-2S2
3)P3=S1P2-S2P1+3S3
4)P4=S1P3-S2P2+S3P1
5)P5=S1P4-S2P3+S3P2
Rest is easy.
Plug in the numbers and let the magic begin
To calculate abc it would have been much easier to use the previous result (a+b+c)^2=2*(1+ab+bc+ac)
So that (a+b+c)^3=(a+b+c)*2*(1+ab+bc+ac)
==>
(a+b+c)+3*abc+(a^2+b^2+c^2)-(a^3+b^3+c^3)=1/2
==>
Abc=1/6
The only man who could defeat Thanos
In a math class we were told that in th xy plane if we have a parallelogram and its base has sqrt(something) then the height of that parallelogram will have the same sqrt(of the same thing)
Ex:
If the base is =2sqrt(13)
Height must have a sqrt(13) in it
Anyone knows why?
Or a proof..!
Thx😊
This is not true....
ice spirit it is true but i do not know why!
Maybe i didn't make it clear but i am 100% sure it is true...
Hight of parallelogram is not dependent on it's base
Thirsty For Knowledge ik but the teacher(a great one) told us that when we have a sqrt of a number in a height of the parallelogram then it must be the same sqrt in the base or it will be a miscalculation in our side and we have to check it
And after so many times his statement is not false a single time but he doesn't tell us why and want us to search for it
@@pholioschenouda5395 What if length of base is 8 units(4√4) and height is 3 units? Now there is no √4 in height.
a^2*b^2+a^2*c^2+b^2*c^2 is negative, and so one has to be complex. But since the sum and product are real, another one has to be the conjugate of that one.
nah, just a consequence of summing all the natural numbers /s
SUS
@@adler4102 yes indeed SUS
the way you arranged the problem in 4:12 reminded me of symmetric matrix diagonalization question. Preety sure its just a coincedence or smth. Great video btw.
This Problem Is Easy to solve.
1) solve for a(1-b-c)
2) Plug in the new vallue for a
3) solve for b
4)Plug in
5) linear equation for c