Much simpler way to do this. x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = 0 So x^3 = -1/x^3 So x^999 = -1/x^999 by raising both sides to 333 So x^999 + 1/x^999 = 0
@@irenehartlmayr8369 you get a different question depending on your subject. You get a question once you do your interview at Oxford, it basically contributes to whether you can make it into the school. For economics for example you'd get a question regarding the economy, choices or something else. If you give a good answer you'd be considered for a place in Oxford so it's essentially like an entrance exam
when you got x^3 = -1/x^3... can't you just power 333 both sides and get x^999 = -1/x^999 and then simply x^999 + 1/x^999 = 0... what is the problem if we do this?? bcz I don't feel any problem doing this
doing power 333 both side make this like this: x^999=-1/x^999 x^999 + 1/x^999 = 0 so this is verified by two trees so this is correct, you found a new way
The complex exponential gives a hint that this can be generalised for x + 1/x = 2 cos(pi/2n) and x^n+1/x^n = 0. I think that the complex view gives quite a nice insight here.
I just made a bruteforce. From the quadratic equation I got x1,2 = (√3±i)/2. Then, according to Moivre's theorem, x1,2 = 1*(cos(π/6) ± i*sin(π/6)) --> (x1,2)^999 = 1*(cos(999π/6) ± i*sin(999π/6)) = cos(π/2) ± i*sin(π/2)) = 0 ± i = ±i. Finally, by substitution, i+(1/i) = 0; (-i) + 1/(-i) = 0. The only answer is 0.
@@brain_station_videos It was related to a problem posed by Sybermath. I began with the general form of it: f(x,a)•f(x,b) = f(x,a+b) + f(x,a-b). Rearranging it to f(x,a+b) = f(x,a)•f(x,b) - f(x,a-b). Now let a = n+1 and b = 1. The next step was to come with a formula for the nᵗʰ row coefficients for f(x,n) where for f(x,1) = c, in terms of powers of c. c c² - 2 c³ - 3c c⁴ - 4c² +2 c⁵ - 5c³ + 5c c⁶ - 6c⁴ + 9c² - 2 etc I figured out the formula and then discovered this was already known, per the On-Line Encyclopedia of Integer Sequences (OEIS). "A034807 Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials." Ha!
It is much quicker to use Euler's formula and note that the absolute value of x is just 1, whereas its argument is either +i*pi/6 or --i*pi/6. Multiply the arguments each by 999 and cancel redundant cycli of i*2*pi, from which it follows that x^999 = x^3 = 1 and x^-999 = x^-3 =-1, or the other way round. In both cases, hence for both roots of x, the sum is zero. Nothing to blow our minds, I suppose!
also when we got x^3 + 1/x^3= 0, can't we just say that x*999 + 1/x^999 = (x^3)^333 + (1/x^3)^333... now as 333 is odd so it must has one factor= x^3 + (1/x)^3 = 0... Hence the whole expression becomes 0... so x^999 + 1/x^999 = 0
@@yashmehta9299 I am not doing this... we had got that x³ + 1/x³ = 0, so x³ =(-1/x³), now we can power 333 on both the sides and we will get- x⁹⁹⁹ =(-1/x⁹⁹⁹), now we will add 1/x⁹⁹⁹ on both the sides to get- x⁹⁹⁹ + 1/x⁹⁹⁹ = 0
(e^(i[pi/6]))^999 Multiply pi/6 by 999, divide 2(pi), now having 333/4, we get 83.25, which means we now have 2(pi)/4, which is pi/2. Now we have e^(i(pi/2)) This will be at the bottom the fraction shown as 1/(x^999). Have e^(i(pi/2)) over 1 to form a fraction and multiply by e^(i(pi/2)) on top and bottom. The pi/2 and pi/2 will combine to make pi. Giving e^(i(pi)) which is -1. Denominators are the same and reduces down to zero. Same process can be used for e^(-i(pi/6)).
I solved it by substituting x (from the quadratic equation)in x^999 + 1/x^999 and by using de Moivre's theorem I got the answer zer0 . After I posted the comment and I saw you revealed another way also 😂😂
It's too easy I did it by another method:- x+1/x= √3 Squaring both sides,we get x²+1/x²=1 Solving this,we get x²=-w,-w² (Where w & w² are cube root of unity) Now putting any value of x² in x⁹⁹⁹+1/x⁹⁹⁹ will give answer
well it's complex.... if u know about complex numbers then i= sqrt(-1) if x=i then x^6= -1 as: i x i = -1 | -1 x i = -i | -i x i = 1 | 1 x i = i using this cyclicity of iota u will get the answer as -1
👍 x+1/x = √3 cubing both sides x^3+1/x^3+3x.1/x(x+1/x) = 3√3 x^3+1/x^3 = 0 x^3 = - 1/x^3 (x^3)^333 = - (1/x^3)^333 x^999 = - 1/x^999 x^999+1/x^999 = 0 other methods - (A) x^2 - √3 x+1 = 0 by quadratic formula x = (√3 +/- i)/2 x = cos π/6 +/- i sin π/6 x = e^(+/- iπ/6) x^999 = e^(+/- 333πi/2) = { e^(2πi) }^{+/- 83) e^(+/- πi/2) = +/- i because e^(2πi) = 1 in both case value of given expression is zero because reciprocal of +/- i is -/+ i (B) x^3+1/x^3 = 0 (x^3)^2 = - 1 x^3 = +/- i x^999 = (+/- i)^333 = +/- i because (+/- i)^332 = 1 as i^4 = 1 in both cases value of given expression is zero. (C) x^3+1/x^3 = 0 now x^999+1/x^999 = (x^3)^333+(1/x^3)^333 if we factorise it one factor will be x^3+1/x^3 hence answer is zero. (D) x = cos π/6 +/- i sin π/6 x^999 = cos { 83(2π)+π/2 } +/- i sin { 83(2π)+π/2 } = +/- i x^999+1/x^999 = 0 (E) by quadratic formula x = (√3 +/- i)/2 x = i(- 1 - i√3)/2 , - i( - 1+i√3)/2 x = iw^2 , - iw x^999 = i^999 , - i^999 because w^3 = 1 x^999 = i^3 , - i^3 because i^996 = (i^4)^249 = 1^249 = 1 x^999 = - i , i value of given expression = 0
Much simpler way to do this.
x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = 0
So x^3 = -1/x^3
So x^999 = -1/x^999 by raising both sides to 333
So x^999 + 1/x^999 = 0
You could also use de'moivre's theorem. The roots are √3/2±i1/2 or z = {cos(π/6) +isin(π/6) }. We know, z^n = cosnx+isinx.
Yea that's exactly how I did it too roots are W/i and its negative
W³=1 and i³=-i.
EXACTLY, bro just like to make it loooonngggg for no reason
Good one, I usualy do it like e^i*π/6, but it is the same thing
That complex solution and 999 power really screams Moivre’s formula
This type of easy questions is asked in government exams in India 😂😂
But still we Indians proudly lag behind in mathematics and innovation. It's not about difficulty
@arsh_arora10 yes because of the education system
Also because of your mentality
Not my mentality but most of the indian students.
well then try to solve JEE ADVANCED MATHEMATICS(NOT of 2024)
Ain't no way oxford has an entrance exam...💀
i think it does
The question is probably from 50 years ago
What are the entrance requirements for Oxford then ?...and not everyone wants,or needs,a degree in mathematics!
Wait till you hear about tmua
@@irenehartlmayr8369 you get a different question depending on your subject. You get a question once you do your interview at Oxford, it basically contributes to whether you can make it into the school.
For economics for example you'd get a question regarding the economy, choices or something else. If you give a good answer you'd be considered for a place in Oxford so it's essentially like an entrance exam
when you got x^3 = -1/x^3... can't you just power 333 both sides and get x^999 = -1/x^999 and then simply x^999 + 1/x^999 = 0... what is the problem if we do this?? bcz I don't feel any problem doing this
doing power 333 both side make this like this:
x^999=-1/x^999
x^999 + 1/x^999 = 0
so this is verified by two trees so this is correct, you found a new way
The complex exponential gives a hint that this can be generalised for x + 1/x = 2 cos(pi/2n) and x^n+1/x^n = 0. I think that the complex view gives quite a nice insight here.
I just made a bruteforce.
From the quadratic equation I got x1,2 = (√3±i)/2.
Then, according to Moivre's theorem, x1,2 = 1*(cos(π/6) ± i*sin(π/6)) --> (x1,2)^999 = 1*(cos(999π/6) ± i*sin(999π/6)) = cos(π/2) ± i*sin(π/2)) = 0 ± i = ±i.
Finally, by substitution, i+(1/i) = 0; (-i) + 1/(-i) = 0. The only answer is 0.
Let f(x,n) = xⁿ + 1/xⁿ, find f(x,999)
f(x,n+2) = f(x,1)•f(x,n+1) - f(x,n) = √3•f(x,n+1) - f(x,n)
f(x,0) = 2
f(x,1) = √3
f(x,2) = 1
f(x,3) = 0 and he sequence repeats every 12 terms
So f(x,999) = f(x,3) = 0
This is good. Btw how you figured out the second step? That recurrence relation..
@@brain_station_videos It was related to a problem posed by Sybermath. I began with the general form of it: f(x,a)•f(x,b) = f(x,a+b) + f(x,a-b). Rearranging it to f(x,a+b) = f(x,a)•f(x,b) - f(x,a-b). Now let a = n+1 and b = 1.
The next step was to come with a formula for the nᵗʰ row coefficients for f(x,n) where for f(x,1) = c, in terms of powers of c.
c
c² - 2
c³ - 3c
c⁴ - 4c² +2
c⁵ - 5c³ + 5c
c⁶ - 6c⁴ + 9c² - 2
etc
I figured out the formula and then discovered this was already known, per the On-Line Encyclopedia of Integer Sequences (OEIS). "A034807 Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials." Ha!
It is much quicker to use Euler's formula and note that the absolute value of x is just 1, whereas its argument is either +i*pi/6 or --i*pi/6. Multiply the arguments each by 999 and cancel redundant cycli of i*2*pi, from which it follows that x^999 = x^3 = 1 and x^-999 = x^-3 =-1, or the other way round. In both cases, hence for both roots of x, the sum is zero. Nothing to blow our minds, I suppose!
you can turn that x value into an exponential and then divide 999 by 12 too man
Didn't expect to be this eazy
also when we got x^3 + 1/x^3= 0, can't we just say that x*999 + 1/x^999 = (x^3)^333 + (1/x^3)^333... now as 333 is odd so it must has one factor= x^3 + (1/x)^3 = 0... Hence the whole expression becomes 0... so x^999 + 1/x^999 = 0
That’s what I thought too
No, because (x^3 + 1/x^3)^333 is not necessarily the same as x^3*333 + 1/x^3*333
@@yashmehta9299 I am not doing this... we had got that x³ + 1/x³ = 0, so x³ =(-1/x³), now we can power 333 on both the sides and we will get- x⁹⁹⁹ =(-1/x⁹⁹⁹), now we will add 1/x⁹⁹⁹ on both the sides to get- x⁹⁹⁹ + 1/x⁹⁹⁹ = 0
@@abacademy8896 ah, lol, yeah
This is literally just square both sides define u = x^2 solve for u find x. Put in original equation and see if it works. Done.
(e^(i[pi/6]))^999
Multiply pi/6 by 999, divide 2(pi), now having 333/4, we get 83.25, which means we now have 2(pi)/4, which is pi/2.
Now we have
e^(i(pi/2))
This will be at the bottom the fraction shown as 1/(x^999).
Have e^(i(pi/2)) over 1 to form a fraction and multiply by e^(i(pi/2)) on top and bottom.
The pi/2 and pi/2 will combine to make pi. Giving e^(i(pi)) which is -1.
Denominators are the same and reduces down to zero. Same process can be used for
e^(-i(pi/6)).
bro pls upload more of there tricky questions more hard with good explanation ,,,,,[in same style]'''''''pls more '''thx
Very nice😮😮😮😮😮😮😮😮
*You were done mid-way (**1:43**) into the video*
If x^3 = -1 / x^3 then x^999 = (x^3)^333 = (-1/x^3)^333 = -1/x^999. Therefore, x^999 + 1/x^999 = 0
Ahh yesss 1 + 1/1 is indeed equal to root of three
😅😅
How to calculate this
"10!"?
n! = n(n-1)(n-2)(n-3)...(3)(2)(1)
Therefore: 10! = 10×9×8×7×6×5×4×3×2×1
Just say “10” loudly
Nice☺☺☺☺☺☺☺
How many ways the three balls can it be arranged when you get it from the group of six balls?
First, we choose 3 balls out of 6 in 6C3 ways and then arrange them in 3! ways
So the ans will be 6C3×3! = 120 👍🏻
Try telling the truth. Unlike you, Oxford wouldn't embarrass itself by making a fuss over such SIMPLE quadratics.
I solved it by substituting x (from the quadratic equation)in x^999 + 1/x^999 and by using de Moivre's theorem I got the answer zer0 .
After I posted the comment and I saw you revealed another way also 😂😂
X,2×+5=8
It's too easy
I did it by another method:-
x+1/x= √3
Squaring both sides,we get
x²+1/x²=1
Solving this,we get x²=-w,-w²
(Where w & w² are cube root of unity)
Now putting any value of x² in x⁹⁹⁹+1/x⁹⁹⁹ will give answer
I don't know if my method is correct,
I done it simply by differentiating separately and using log formula I got 0 as answer
Bro we can manipulate root as cube root of unity 🫣
make videos like ( or animation, to be more specific ) like 3blue1brown.
He has his own style, and I respect that and also love it.
Why make everything so complex when this itself is super easy to follow
And it's also the fact that if he does this he'd get hate for exactly that, reminds me of all "Mrbeast" type of videos
Tried avoiding the value of x to no avail
It can be changed to a cube root of unity so…. Yeah
Yes we can solve by euler formula🎉
here x is - of complex no. W so W⁹⁹⁹ =1 so ans is 2
I am an eight standard student . I solved this problem within 2 min .😂
Pls I can't handle sin, cos, tan, pi, and i
And therefore i have solved it without using them
Let .
Step 1: Express
Using the exponential form of , we have:
x = e^{i\pi/6}.
x^{999} = \left(e^{i\pi/6}
ight)^{999} = e^{i \cdot 999 \cdot \pi / 6}.
Simplify the exponent:
999 \cdot \frac{\pi}{6} = 166.5\pi = 166\pi + 0.5\pi.
166.5\pi \mod 2\pi = 0.5\pi.
x^{999} = e^{i \cdot 0.5\pi} = i.
---
Step 2: Find
The reciprocal of is:
\frac{1}{x^{999}} = \frac{1}{i} = -i.
---
Step 3: Compute
Now, substitute:
x^{999} + \frac{1}{x^{999}} = i + (-i) = 0.
---
Final Answer:
x^{999} + \frac{1}{x^{999}} = 0.
chatgpt momento, haha ha h
Good 👏
def chatgpt because of the "\frac{1}{69}\" format on fractions
Please can someone explain me how x^6 can be negative?
Have u heard about complex number
well it's complex....
if u know about complex numbers then i= sqrt(-1)
if x=i then x^6= -1
as: i x i = -1 | -1 x i = -i | -i x i = 1 | 1 x i = i
using this cyclicity of iota u will get the answer as -1
@@Rishith198 thanks for the explanation though
@@Alphamatics1234 I don’t know anything about them
A simple one you got equation as x³=-1/x³ so multiply both sides by x³ you will get --> x⁶=-x³/x³. Cancel out x³ on RHS to get -1
How did we manage to turn math into brainrot
{x+x ➖}+{1+1 ➖ }/{x+x ➖}={x^2+2}/x^2=2x^2/x^2=2x^1 (x ➖ 2x+1).6 3^3 (x ➖ 3x+3). {x^999+x^999 ➖}+{1+1 ➖ }/{x^999+x^999➖ }={x^1998+2}/x^1998=2x^1998/x^1998= 2x^3^2^3^2^2^3/x^3^2^3^2^2^3 2x^1^1^1^1^1^1^1/x^1^1^1^1^1^1^1 2x^1/x^1 2x^1/ (x ➖ 2x+1).
i am an indian(asian) so this is easy for me
Hello bro I am also indian. 😁
Tumhara naam kya hai
But I am in 10th grade.
Aur mai Haryana se hu
@@rohitkaushik2311 mai 9 grade me hu
Ohhhh,my Head
Omg!The answer may be zero.......
Based on the key x^6=(-1)...(^=read as to the power )
Explain later
First equation is quadratic so you can solve in memory. Then substitute in second
Very easy
👍
x+1/x = √3
cubing both sides
x^3+1/x^3+3x.1/x(x+1/x) = 3√3
x^3+1/x^3 = 0
x^3 = - 1/x^3
(x^3)^333 = - (1/x^3)^333
x^999 = - 1/x^999
x^999+1/x^999 = 0
other methods -
(A)
x^2 - √3 x+1 = 0
by quadratic formula
x = (√3 +/- i)/2
x = cos π/6 +/- i sin π/6
x = e^(+/- iπ/6)
x^999 = e^(+/- 333πi/2)
= { e^(2πi) }^{+/- 83) e^(+/- πi/2)
= +/- i because e^(2πi) = 1
in both case value of given expression is zero because reciprocal of +/- i is -/+ i
(B)
x^3+1/x^3 = 0
(x^3)^2 = - 1
x^3 = +/- i
x^999 = (+/- i)^333 = +/- i
because (+/- i)^332 = 1 as i^4 = 1
in both cases value of given expression is zero.
(C)
x^3+1/x^3 = 0
now x^999+1/x^999
= (x^3)^333+(1/x^3)^333
if we factorise it one factor will be x^3+1/x^3 hence answer is zero.
(D)
x = cos π/6 +/- i sin π/6
x^999 = cos { 83(2π)+π/2 } +/- i sin { 83(2π)+π/2 } = +/- i
x^999+1/x^999 = 0
(E)
by quadratic formula
x = (√3 +/- i)/2
x = i(- 1 - i√3)/2 , - i( - 1+i√3)/2
x = iw^2 , - iw
x^999 = i^999 , - i^999 because w^3 = 1
x^999 = i^3 , - i^3 because i^996 = (i^4)^249 = 1^249 = 1
x^999 = - i , i
value of given expression = 0
Amazing
Hmm ... this question was a little bit of no. Theory good good 👍🏻
oh hell no
just use de moivre