Extreme Algebra Question (

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  • Опубліковано 24 гру 2024

КОМЕНТАРІ • 2,2 тис.

  • @blackpenredpen
    @blackpenredpen  5 років тому +3494

    EXTREME CHALLENGE:
    Find a, b, and c individually!

  • @Andrew..J
    @Andrew..J 5 років тому +2894

    I'm just gonna say 5 and hope the rest of the test is easier.

    • @maxwell_abe2894
      @maxwell_abe2894 4 роки тому +36

      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

    • @Mukawakadoodoo
      @Mukawakadoodoo 3 роки тому +25

      What if it’s the only question on the test 🤣🤣🤣

    • @Me-vu3yv
      @Me-vu3yv 3 роки тому +15

      @@Mukawakadoodoo then you are in trouble

    • @naturalaviator4156
      @naturalaviator4156 3 роки тому +2

      Lmao

    • @parthibhayat
      @parthibhayat 3 роки тому

      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

  • @rob6129
    @rob6129 5 років тому +5332

    hes becoming a real professor: occasionally confused by himself

    • @blackpenredpen
      @blackpenredpen  5 років тому +411

      R Hahahhahahaahahaha right on!

    • @mystudy6827
      @mystudy6827 5 років тому +29

      Hes becoming a real professor:

    • @Pygmygerbil88
      @Pygmygerbil88 5 років тому +10

      @@mystudy6827 isn he?

    • @LORD-kg4sl
      @LORD-kg4sl 4 роки тому +18

      He is already a teacher lmao

    • @nickpapagiorgio9872
      @nickpapagiorgio9872 4 роки тому +7

      LORD hopefully he learned to slow down so people can understand . It’s not a race.

  • @darionsama6612
    @darionsama6612 5 років тому +2943

    *pauses the video to try to solve it*
    *notices the 41 minutes length*
    *plays the video and throws the notebook away*

    • @hrutikabadgeri6505
      @hrutikabadgeri6505 5 років тому +22

      😂😂😂

    • @NovaWarrior77
      @NovaWarrior77 5 років тому +42

      I applaud you! Glad I was not the only one who tried to solve it a little.

    • @oximas
      @oximas 5 років тому +1

      i triend to solve it then realease it would take for ever to solve the problem and i am getting nothing afterwards

    • @rishabhbharti3580
      @rishabhbharti3580 4 роки тому +1

      🤣🤣🤣🤣🤣🤣🤣🤣

    • @ashumishra7020
      @ashumishra7020 4 роки тому +5

      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

  • @mustafakhurramaziz9715
    @mustafakhurramaziz9715 4 роки тому +488

    Me:*looks at whiteboard*
    Whiteboard: 0:30
    Me:*looks outside the window for 0.000001seconds
    The whiteboard: 40:04

    • @XoIoRouge
      @XoIoRouge 3 роки тому +21

      It's a shame the joke has to use timestamps. Wish it could've used pictures. 😂

    • @deleted-something
      @deleted-something 3 роки тому +2

      True

    • @exzar9684
      @exzar9684 2 роки тому +1

      in class everytime 🤣🤣🤣

  • @blazingfire7517
    @blazingfire7517 5 років тому +185

    “Try this first”
    Me seeing that this is a 40 minute video: “no, I don’t think I will”

  • @LuisMartinez-ih2oc
    @LuisMartinez-ih2oc 5 років тому +5972

    Teachers example:
    4x-3y=6
    y=-3x+15
    Classwork:
    x-3y+3z=-4
    2x+3y-z=15
    4x-3y-z=19
    Exam:

    • @ilya9481
      @ilya9481 5 років тому +514

      Exam: prove a^n + b^n ≠ c^n (n>2)

    • @justaconcrete4789
      @justaconcrete4789 5 років тому +164

      @@ilya9481 Hahah that's the last theorem of Fermat's, Incredible brain to solve it

    • @adityashankar5723
      @adityashankar5723 5 років тому +36

      @@ilya9481
      I actually started solving that.

    • @GodOnVayne
      @GodOnVayne 5 років тому +23

      (5/3, 44/9, 3) (x,y,z)

    • @bencroshaw7293
      @bencroshaw7293 5 років тому +22

      @@GodOnVayne . Did you use matrix multiplication to work the equation out?

  • @hatasesasd9631
    @hatasesasd9631 5 років тому +472

    15:55 * Erases K *... No, It's C to the k power * Writes a new K *

    • @ngochoang4639
      @ngochoang4639 3 роки тому +2

      Lol

    • @BlankTH
      @BlankTH 3 роки тому +23

      lmao ye and the way he goes "whew" afterwards like thank goodness I noticed. omg my sides

    • @arah8998
      @arah8998 3 роки тому +22

      He fixed it in minute 20 (20:00)

    • @Nio10895
      @Nio10895 3 роки тому +2

      Thats so me

    • @albusdumbledore2406
      @albusdumbledore2406 3 роки тому +2

      That ruined my ocd

  • @andyb6177
    @andyb6177 3 роки тому +184

    "Try it for yourself"
    *opens MatLab*
    *gets answer with less then 10 lines of code*

    • @beechass4451
      @beechass4451 3 роки тому +4

      B-b-but that's not the point man :(

    • @78anurag
      @78anurag 3 роки тому +2

      Code pls

    • @darrylchandra554
      @darrylchandra554 3 роки тому +1

      But TBH it ain't hard, it is just complicated

    • @AryanSingh-yb2my
      @AryanSingh-yb2my 3 роки тому +3

      How did you write the code, could you share yours, ive been trying to make it but I just cant:( send yours pls

  • @cblpu5575
    @cblpu5575 2 роки тому +10

    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
      @savaesmek 2 роки тому

      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

    • @cblpu5575
      @cblpu5575 2 роки тому

      @@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.

    • @savaesmek
      @savaesmek 2 роки тому

      @@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
      @cblpu5575 2 роки тому

      @@savaesmek haha, happens all the time

    • @savaesmek
      @savaesmek 2 роки тому

      @@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

  • @blackpenredpen
    @blackpenredpen  5 років тому +709

    Mistake at 3:06 . a^2*c^3 should be a^3*c^2 but the rest is okay.

    • @vitakyo982
      @vitakyo982 5 років тому +19

      I was going to comment , but ...

    • @leadeer4213
      @leadeer4213 5 років тому +33

      15:50 u wrote k no c in base

    • @omarmachuca6405
      @omarmachuca6405 5 років тому +7

      blackpenredpen How is this written 2 days ago?

    • @leadeer4213
      @leadeer4213 5 років тому +1

      @@omarmachuca6405 That video wasn't public

    • @sarthakgarg5701
      @sarthakgarg5701 5 років тому +1

      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

  • @Nickesponja
    @Nickesponja 5 років тому +2373

    If I ever become a college professor, I'm going to put this as a common exercise and tell my students it's trivial

    • @General12th
      @General12th 5 років тому +266

      "The proof is an exercise for the reader."

    • @jahad6335
      @jahad6335 5 років тому +51

      But... it is actually trivial

    • @General12th
      @General12th 5 років тому +273

      @@jahad6335 If the derivation takes forty minutes, it's not trivial. You're just trying to make yourself look smart. Cut it out.

    • @Nickesponja
      @Nickesponja 5 років тому +92

      @@General12th One time a professor told me that proofs start being not trivial when they take more than 3 pages xD

    • @alessandrofenu4325
      @alessandrofenu4325 5 років тому +56

      @@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).

  • @pleasepassthemustard4401
    @pleasepassthemustard4401 5 років тому +363

    That one kid in math class who does this even when the teacher says you don't have to show your work

  • @wanlitan7406
    @wanlitan7406 4 роки тому +91

    0:14 "We know the fourth power is going to be too easy"
    Uh, excuse me?

    • @nilarghyachatterjee5392
      @nilarghyachatterjee5392 Місяць тому

      yea it came in one of my exams and i took 10 minutes solving it 😭( i am in 6)

  • @505steel
    @505steel 5 років тому +64

    I love how he is always so excited about math! Such an inspiration

  • @toopytoopy8547
    @toopytoopy8547 5 років тому +332

    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
      @boong0006 5 років тому +2

      Thank you!

    • @boong0006
      @boong0006 5 років тому +6

      Since P(X) is a trinomial, I wonder if you can apply this method to a quadnomial that involves a,b,c and d?

    • @dacres2002
      @dacres2002 4 роки тому +6

      Thats the best comment ever!!

    • @dacres2002
      @dacres2002 4 роки тому +9

      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

    • @toopytoopy8547
      @toopytoopy8547 4 роки тому +16

      @@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.

  • @sinzuii2061
    @sinzuii2061 5 років тому +672

    Skipping from 0:00 to 28:00
    Is like my teacher after i took a restroom.

    • @WilliMel
      @WilliMel 5 років тому +33

      @@codeguy21 dafaq

    • @ifteharulhaque2600
      @ifteharulhaque2600 5 років тому +1

      @@codeguy21 wut?

    • @fdnt7_
      @fdnt7_ 5 років тому +1

      @@codeguy21 TF?

    • @molozu6370
      @molozu6370 5 років тому +24

      damn bro pass me a restroom too

    • @jiauddin007
      @jiauddin007 4 роки тому

      that means you go to the restroom for 28 minutes.

  • @micronalpha
    @micronalpha 5 років тому +234

    Congrats on showing the non-edited video. We've all been there, batteling hard algebra. I enjoyed seeing this video. ☺

    • @blackpenredpen
      @blackpenredpen  5 років тому +23

      Mauro Maia thank you!!!!

    • @anshumanagrawal346
      @anshumanagrawal346 5 років тому +2

      @@blackpenredpen I found a way to solve this problem which is very easy and short

    • @dux2508
      @dux2508 5 років тому +1

      Anshuman Agrawal Please share

    • @MrKrabs-xf2tr
      @MrKrabs-xf2tr 3 роки тому +1

      @@dux2508 Simply use Newton's Identities and consider a,b, and c as roots of a cubic

    • @anshumanagrawal346
      @anshumanagrawal346 2 роки тому

      @@MrKrabs-xf2tr Yes, exactly

  • @raphielohnef4678
    @raphielohnef4678 5 років тому +3

    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.

  • @JonathonV
    @JonathonV 4 роки тому +15

    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.

  • @vibhavaggarwal237
    @vibhavaggarwal237 5 років тому +634

    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.

    • @peterderias5323
      @peterderias5323 5 років тому +63

      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

    • @vibhavaggarwal237
      @vibhavaggarwal237 5 років тому +9

      @@peterderias5323 That's interesting :)

    • @gekkouga5147
      @gekkouga5147 5 років тому +15

      I didn't get it😥

    • @vibhavaggarwal237
      @vibhavaggarwal237 5 років тому +31

      @@gekkouga5147 Have you studied polynomials? If not, I would suggest to study that first. If yes, tell me which part is troubling you.

    • @SonGoku-rx1fo
      @SonGoku-rx1fo 5 років тому +12

      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 😅.

  • @JoshuaHillerup
    @JoshuaHillerup 5 років тому +789

    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.

    • @blackpenredpen
      @blackpenredpen  5 років тому +167

      Joshua Hillerup hahahhahahahha

    • @richardbembridge1822
      @richardbembridge1822 5 років тому +37

      Brilliant.

    • @1remphase1
      @1remphase1 5 років тому +11

      Winner 🙈

    • @fourier07able
      @fourier07able 5 років тому +24

      For sure they change your symbols by a, b, c, or x, y, z, etc.

    • @VENOM-tx6gp
      @VENOM-tx6gp 5 років тому +268

      Wait you MUST NOT forget to write 98% fails to solve

  • @stickman32032
    @stickman32032 5 років тому +104

    Proud to admit I watched the video all the way through instead of doing my homework

  • @nicholaswongso2003
    @nicholaswongso2003 3 роки тому +41

    bprp: "find a, b, and c individually!"
    me: *no*

  • @georgemissailidis1504
    @georgemissailidis1504 5 років тому +47

    And what sucks is that if you make *one* mistake, everything to follow is all wrong.

    • @NavsangeetSingh
      @NavsangeetSingh 4 роки тому +11

      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.

    • @nocturnalvisionmusic
      @nocturnalvisionmusic 4 місяці тому

      That's actually a good comment George :)

  • @Arbmosal
    @Arbmosal 5 років тому +7

    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.

  • @goup867
    @goup867 5 років тому +11

    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.

    • @davidfranklin2731
      @davidfranklin2731 5 років тому

      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).

    • @ayanjain3106
      @ayanjain3106 5 років тому

      I too did with the same method .

    • @goup867
      @goup867 5 років тому

      @Cool Dude I am from 10th grade too so they probably may :)

  • @hunghinsun2123
    @hunghinsun2123 5 років тому +8

    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.

    • @nilsh5027
      @nilsh5027 3 роки тому

      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

  • @ethandavies953
    @ethandavies953 3 роки тому +1

    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

  • @somenamelastnaammee52
    @somenamelastnaammee52 3 роки тому +1

    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

  • @ffggddss
    @ffggddss 5 років тому +6

    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
      @blackpenredpen  5 років тому +2

      Sorry for the ads. This video is longer than usual.

    • @ffggddss
      @ffggddss 5 років тому +2

      @@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

    • @M-F-H
      @M-F-H 4 роки тому +1

      Hi Fred. To avoid the ads, go to the end of the video and then do "replay".

    • @ffggddss
      @ffggddss 4 роки тому

      @@M-F-H Thanks! I'll try that.
      Fred

  • @benhetland576
    @benhetland576 5 років тому +12

    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? :-)

  • @Debg91
    @Debg91 5 років тому +8

    Awesome! Your exercises are getting more and more interesting!

  • @싱미-h2f
    @싱미-h2f 2 роки тому +3

    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

  • @backtothechickens
    @backtothechickens 4 роки тому +7

    Opens video with "lets do an extereme algebra question for fun". This is the youth culture I have been looking for.

  • @Andy-ju8bb
    @Andy-ju8bb 5 років тому +6

    Just watched my 3rd back-to-back blackpenredpen video. My brain now needs a vacation on a tropical island.

  • @veetee355
    @veetee355 5 років тому +539

    What the hell is this.

    • @jasonstarrising
      @jasonstarrising 4 роки тому +2

      vtt355
      It’s algebra 2 but like the accelerated version.

    • @gregoriousmaths266
      @gregoriousmaths266 4 роки тому

      Craziness

    • @AZ-ps1si
      @AZ-ps1si 4 роки тому

      Our Worst dream.

    • @R97-c5h
      @R97-c5h 4 роки тому

      Abdulaziz Alquhaibi عبدالعزيز هذا شنو 😂

    • @nameless5678
      @nameless5678 4 роки тому

      A nightmare

  • @isabahk1132
    @isabahk1132 5 років тому +61

    *If this was my exam, it would just be worth like 6 marks, nothing else*

  • @rashmigupta6227
    @rashmigupta6227 4 роки тому +1

    Very easy if you know
    a^3+b^3+c^3 -3abc=(a+b+c)(a^ 2b^2 c^2.-ab -bc -ca )

  • @disgruntledtoons
    @disgruntledtoons 2 роки тому

    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).

  • @duggydo
    @duggydo 5 років тому +67

    38:00 implies complex solutions! 😄

    • @blackpenredpen
      @blackpenredpen  5 років тому +4

      duggydo pretty much!

    • @typo691
      @typo691 5 років тому

      What makes you say so? Experience or memes?

    • @blackpenredpen
      @blackpenredpen  5 років тому +3

      Typo ???

    • @reetasingh1679
      @reetasingh1679 5 років тому +14

      @@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.

    • @keescanalfp5143
      @keescanalfp5143 5 років тому

      @duggydo, From the first minute we thought of a kind of symmetrical couple, triple, in the complex plane.

  • @auroremelie
    @auroremelie 5 років тому +321

    You should invest into a small microphone that you can attach to your clothe.

  • @jonathandavis8014
    @jonathandavis8014 3 роки тому +17

    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!

  • @aryam9134
    @aryam9134 4 роки тому +1

    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!
    :)

  • @ibbeeabro2288
    @ibbeeabro2288 3 роки тому +1

    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)

  • @davidnobre171
    @davidnobre171 5 років тому +137

    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

    • @milindbordia
      @milindbordia 5 років тому +7

      David Nobre don’t worry, time is relative

    • @x15cyberrush9
      @x15cyberrush9 5 років тому

      hahaha

    • @crosisbh1451
      @crosisbh1451 5 років тому +10

      That's when I inconspicuously add/drop a negative somewhere in my work to just make it work out.

    • @keescanalfp5143
      @keescanalfp5143 5 років тому +3

      @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.
      **

    • @ffggddss
      @ffggddss 5 років тому +5

      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

  • @Catishcat
    @Catishcat 5 років тому +40

    ...
    I only realised I just watched 40 minutes of algebra after the video finished
    Math is immersive. Can't believe it was 40 minutes...

    • @blackpenredpen
      @blackpenredpen  5 років тому +2

      : ))))))

    • @bardozan
      @bardozan 5 років тому

      Can't believe it was only 40 minutes. Would watch for a couple of hours.

    • @jofx4051
      @jofx4051 5 років тому +1

      20 minutes if you 2X it

  • @avdeshkumar8500
    @avdeshkumar8500 5 років тому +69

    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

  • @AdityaKumar-gv4dj
    @AdityaKumar-gv4dj 2 роки тому

    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

  • @UnbreakablePickaxe
    @UnbreakablePickaxe 3 роки тому +2

    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))

    • @alejofossati
      @alejofossati 3 роки тому

      Idk if you are right but congrats for even knowing what he's talking about.

  • @stevehof
    @stevehof 5 років тому +4

    I love this guy! He makes math more entertaining than a lot of Hollywood movies I've seen lately!

  • @alejandromahillo3536
    @alejandromahillo3536 5 років тому +6

    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 ^^

    • @AnuragGuptainspired
      @AnuragGuptainspired 4 роки тому

      How can i save your solution?😊

    • @DD-rl7xo
      @DD-rl7xo 4 роки тому

      @@AnuragGuptainspired Take a screenshot

  • @bot831
    @bot831 5 років тому +8

    imagine that the camera werent recording

  • @keshavrai4870
    @keshavrai4870 3 роки тому +1

    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

  • @cmilkau
    @cmilkau 4 роки тому

    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.

  • @appanrakaraddi9120
    @appanrakaraddi9120 5 років тому +5

    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.

    • @jrajesh11
      @jrajesh11 5 років тому

      That’s what I thought too!

  • @ssdd9911
    @ssdd9911 5 років тому +74

    using the same mathod, a^4+b^4+c^4=25/6

    • @matteodamiano6733
      @matteodamiano6733 5 років тому +23

      Thats almost 4

    • @ammarbarbhaiwala9908
      @ammarbarbhaiwala9908 5 років тому +5

      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.

    • @fiNitEarth
      @fiNitEarth 5 років тому +2

      @@ammarbarbhaiwala9908 wie zum fick kommt einer Mit mois Profilbild unter so ein mathe Video? XD

    • @AndDiracisHisProphet
      @AndDiracisHisProphet 5 років тому

      @@fiNitEarth Was für ein Typ? Das ist JC Denton

    • @badereric
      @badereric 5 років тому

      @@AndDiracisHisProphet there is no god

  • @BrainGainzOfficial
    @BrainGainzOfficial 5 років тому +10

    Patience is key! Love these kind of videos man, keep it up!

  • @SyberMath
    @SyberMath 3 роки тому

    14:36 I love the Trinomial Expansion!!!
    Cool solution 🤩

  • @gurkiratsingh7tha993
    @gurkiratsingh7tha993 3 роки тому

    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.

  • @aav56
    @aav56 3 роки тому +12

    "Who tf would watch a 40 minute video of a guy doing algebra???"
    -Me, 40 minutes ago

  • @x15cyberrush9
    @x15cyberrush9 5 років тому +164

    i thought answer would be -1/12

    • @keescanalfp5143
      @keescanalfp5143 5 років тому +2

      @X15 CYBER RUSH,
      Just try
      a¹² + b¹² + c¹²
      You will be surprised.
      We were too.
      Good luck!

    • @uddeshyaraj2848
      @uddeshyaraj2848 5 років тому +2

      @@keescanalfp5143 how did u write 12 in superscript

    • @keescanalfp5143
      @keescanalfp5143 5 років тому +5

      @@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?

    • @phil_lx
      @phil_lx 5 років тому +1

      @@uddeshyaraj2848 what about using the app Unicode pad

    • @ar3568row
      @ar3568row 4 роки тому

      @@uddeshyaraj2848 hold the 1 and 2

  • @deusexmaximum8930
    @deusexmaximum8930 5 років тому +29

    "Please pause the video and try this first."
    Oh that's alright I'm alrighty finished

  • @aayushmishra130
    @aayushmishra130 2 роки тому

    you can do it by newton sum by assuming a,b,c as roots of a cubic polynomial

  • @kylekyle3305
    @kylekyle3305 4 роки тому +1

    Don’t stress easy way:9a^3> and now convert it down

  • @togbot3984
    @togbot3984 5 років тому +13

    8:16 that's literally me. I do always solve algebra and get a unexpected result.

  • @shailesh_rajpurohit
    @shailesh_rajpurohit 4 роки тому +8

    16:00 he again write k in place of c😅😅😁

  • @piept4651
    @piept4651 5 років тому +9

    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.

    • @blackpenredpen
      @blackpenredpen  5 років тому +4

      : )))
      I didn't want to do that since (a+b+c)^5 is going to be crazy

  • @ArchanaSharma-eh2it
    @ArchanaSharma-eh2it 5 років тому +1

    If it just to find , you can easily do it by creating a cubic whose roots are a , b , and c

  • @nikhilnagaria2672
    @nikhilnagaria2672 3 роки тому

    I remember watching it about 2 years back. Man, how much have I grown, how much has changed from then

  • @Mau365PP
    @Mau365PP 5 років тому +32

    Ok now solve for a,b and c

    • @michaelz2270
      @michaelz2270 5 років тому +13

      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 ;)

    • @incription
      @incription 5 років тому

      I thought that's what was being shown, but was dissapointed :( is it hard to simultaniously solve them?

    • @Susanmugen
      @Susanmugen 5 років тому +3

      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.

    • @incription
      @incription 5 років тому +4

      believe in math, not wolfram alpha!

    • @michaelz2270
      @michaelz2270 5 років тому +12

      @@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.

  • @heribertobarahona7695
    @heribertobarahona7695 5 років тому +11

    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.

  • @user9287p
    @user9287p 5 років тому +13

    Every Algebra person: Yeah I can do Algebra it's super easy I can do it in my head.
    BlackpenRedpen: Hold my beer

  • @nol2521
    @nol2521 3 роки тому +1

    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

  • @prajwaldeepkamble6617
    @prajwaldeepkamble6617 4 роки тому +1

    3b1b for visualizing maths.
    BpRp for learning maths.
    Thats more than enough.

  • @virakboththan3550
    @virakboththan3550 5 років тому +6

    This is so amazing I’m so surprised

  • @mohdaftab2819
    @mohdaftab2819 4 роки тому +7

    I tried this question with newton's identities by making a polynomial with roots a b and c and got the right answer

    • @nuzhatjahan4781
      @nuzhatjahan4781 3 роки тому

      That's what I was looking for

    • @anmoljhamb9190
      @anmoljhamb9190 3 роки тому +1

      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!

  • @yogipandit9086
    @yogipandit9086 4 роки тому +3

    I wanna know how (ab)^2 + (bc)^2 + (ca)^2 is -1/12

    • @ΜηδὲνἌγαν-β7φ
      @ΜηδὲνἌγαν-β7φ 3 роки тому

      Exactly! The solution is fake. Because a sum of positives makes an negative! This means that the supposes at the begining is impossible!

    • @anshumanagrawal346
      @anshumanagrawal346 3 роки тому

      @@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

  • @yasharya9765
    @yasharya9765 4 роки тому +1

    Better method can be by writing recurrence relation

  • @jrajesh11
    @jrajesh11 5 років тому +1

    Brilliant working out impromptu on board. Really good!!

  • @ssdd9911
    @ssdd9911 5 років тому +546

    -1/12!

  • @TimesOfSilence
    @TimesOfSilence 5 років тому +27

    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.

    • @chrissekely
      @chrissekely 5 років тому +18

      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.

    • @t_kon
      @t_kon 5 років тому +7

      No it'll be messy

    • @Nickesponja
      @Nickesponja 5 років тому +13

      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...

    • @jkn6644
      @jkn6644 5 років тому

      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.

    • @Nickesponja
      @Nickesponja 5 років тому

      @@jkn6644 solve it using Cardan's method (or put it on Wolfram alpha)

  • @edwardhuff4727
    @edwardhuff4727 5 років тому +7

    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.

  • @gvssen
    @gvssen 4 роки тому

    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

  • @satyapalsingh4429
    @satyapalsingh4429 4 роки тому

    Very Nice Method Of Teaching .Stay Blessed .

  • @xyrokai6343
    @xyrokai6343 4 роки тому +71

    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*

    • @blackpenredpen
      @blackpenredpen  4 роки тому +9

      Lolllll

    • @xyrokai6343
      @xyrokai6343 4 роки тому +18

      @@blackpenredpen Fun fact İ was trying to solve this in my dream. Thank you for making it a nightmare.

    • @DJ-mr9tg
      @DJ-mr9tg 4 роки тому

      When you realise that actually isn't a joke...

    • @matematik-ks3860
      @matematik-ks3860 4 роки тому

      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

  • @studyintothefuture3142
    @studyintothefuture3142 4 роки тому +8

    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!

  • @fanamatakecick97
    @fanamatakecick97 4 роки тому +7

    I’ve never heard him swear before. That was hilarious

  • @ConstantUNTILisnt
    @ConstantUNTILisnt 3 роки тому +1

    In the first observation, 3rd line, instead of (a^2)(c^3) isnt it supposed to be (a^3)(c^2)?

  • @girishkumar-xs9go
    @girishkumar-xs9go 3 роки тому

    You can use a³+b³+c³=(a+b+c)³-3(a+b)(b+c)(c+a) to get value of (abc).

  • @Hexanitrobenzene
    @Hexanitrobenzene 5 років тому +4

    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 :) )

  • @petterhouting7484
    @petterhouting7484 5 років тому +4

    You know its a fun question when even bprp gets confused

  • @eizwaneiz
    @eizwaneiz 4 роки тому +7

    How many times you wanna be confused with your own answer ?
    him: yes. eh? wait..oh yeah yes. it's yes.

  • @salmanpary4723
    @salmanpary4723 4 роки тому +1

    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.

    • @salmanpary4723
      @salmanpary4723 4 роки тому

      Plug in the numbers and let the magic begin

  • @antonyqueen6512
    @antonyqueen6512 3 роки тому

    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

  • @raulvalgar1999
    @raulvalgar1999 5 років тому +11

    The only man who could defeat Thanos

  • @pholioschenouda5395
    @pholioschenouda5395 5 років тому +4

    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😊

    • @icespirit
      @icespirit 5 років тому

      This is not true....

    • @pholioschenouda5395
      @pholioschenouda5395 5 років тому

      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...

    • @hmm1778
      @hmm1778 5 років тому

      Hight of parallelogram is not dependent on it's base

    • @pholioschenouda5395
      @pholioschenouda5395 5 років тому

      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

    • @hmm1778
      @hmm1778 5 років тому

      @@pholioschenouda5395 What if length of base is 8 units(4√4) and height is 3 units? Now there is no √4 in height.

  • @stilljust-me2795
    @stilljust-me2795 5 років тому +14

    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.

    • @OwlRTA
      @OwlRTA 3 роки тому

      nah, just a consequence of summing all the natural numbers /s

    • @adler4102
      @adler4102 3 роки тому

      SUS

    • @tslegendary7311
      @tslegendary7311 3 роки тому

      @@adler4102 yes indeed SUS

  • @aissofoxad
    @aissofoxad 3 роки тому

    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.

  • @epicmorphism2240
    @epicmorphism2240 5 років тому

    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