Solving a Cubic equation Using an Algebraic Trick

Поділитися
Вставка
  • Опубліковано 30 жов 2020
  • Join this channel to get access to perks:→ bit.ly/3cBgfR1
    My merch → teespring.com/stores/sybermat...
    Follow me → / sybermath
    Subscribe → ua-cam.com/users/SyberMath?sub...
    Suggest → forms.gle/A5bGhTyZqYw937W58
    If you need to post a picture of your solution or idea:
    intent/tweet?text...
    #ChallengingMathProblems #PolynomialEquations
    EXPLORE 😎:
    Simplifying a Complicated Algebraic Expression: • Simplifying a Complica...
    Solving a Polynomial System in Two Ways: • Solving a Polynomial S...
    A Trigonometry Challenge: • A Trigonometry Challenge
    PLAYLISTS 🎵 :
    Number Theory Problems: • Number Theory Problems
    Challenging Math Problems: • Challenging Math Problems
    Trigonometry Problems: • Trigonometry Problems
    Diophantine Equations and Systems: • Diophantine Equations ...
    Calculus: • Calculus

КОМЕНТАРІ • 107

  • @nilsastrup8907
    @nilsastrup8907 3 роки тому +55

    I love how you explain how to think in these challenging problems in stead of just doing everything correct right away, because many solutions can involve steps that most people wouldn't come up with themselves.

    • @SyberMath
      @SyberMath  3 роки тому +8

      Thank you!

    • @leif1075
      @leif1075 2 роки тому

      That's depressing why do you say most ppl woukdnt come up with themselves..are most ppl not very smart then?

    • @jonathansobieski2962
      @jonathansobieski2962 Рік тому +1

      Coming up with something like this in your own is something only extremely smart people are likely to figure out.

  • @damiennortier8942
    @damiennortier8942 2 роки тому +10

    To get all the solution, when you have 2x^3 - (x-2)^3=0, you can factor out by using a^3 - b^3 = (a-b)(a^2 + ab + b^2) and we know that a = crt2*x and b = x-2 so we can substitute : (crt2*x-x+2)((crt2*x)^2 -crt2*x(x-2) + (x-2)^2) = 0 => [(crt2 - 1)x + 2][(crt2)^2*x^2 - crt2*x^2 + 2crt2*x + x^2 - 4x + 4] = 0 -> [(crt2 - 1)x + 2][(crt2^2 - crt2 + 1)*x^2 + (2crt2 - 4)*x + 4] = 0. And then, we get (crt2 - 1)x + 2 = 0 or (crt2^2 - crt2 +1)x^2 + (2crt2 - 4)x + 4 = 0
    So for the first equation, we get x = - 2/(crt2-1) and for second, delta = 4*crt2^2 - 16*crt2 + 16 - 16crt2^2 + 16crt2 - 16 = - 12*crt2^2 so x1= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x2 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
    So the answer are x1 = - 2/(crt2-1), x2= - [crt2^2 - crt2 + 1 + i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2) and x3 = - [crt2^2 - crt2 + 1 - i*sqrt12*crt2] over (2crt2^2 - 2crt2 + 2)
    Crt = cuberoot and sqrt = squareroot

  • @alnitaka
    @alnitaka 3 роки тому +11

    At 8:40 you take the cube root of both sides. Let w = (-1+sqrt(-3))/2. We can conclude cbrt(2) = x-2, but we could also conclude w*cbrt(2)=x-2, as well as w^2*cbrt(2) = x-2, and this gives the other two roots.

    • @SyberMath
      @SyberMath  3 роки тому +2

      Nice!

    • @alnitaka
      @alnitaka 2 роки тому

      @Arthur So the idea is to only put positive numbers in a square root sign. Seems good to me. Since the square root symbol means the positive root, that should help.

  • @misterdubity3073
    @misterdubity3073 3 роки тому +3

    Teaching people the thought process to search for a solution. Very good!

  • @leonhardeuler5211
    @leonhardeuler5211 3 роки тому +22

    This is smart!

  • @esteger1
    @esteger1 2 роки тому +1

    Nice trick! It was the third one I tried. Finding the complex roots isn't too hard, but writing them on the screen would make everyone's eyes glaze over.

  • @mathsfamily6766
    @mathsfamily6766 3 роки тому +1

    That’s is so amazing trick and clearly how to do it ,great explaining!

  • @golddddus
    @golddddus 3 роки тому +15

    8:18 There are two more solutions. 2x^3-(x-2)3=[2^(1/3)x-(x-2)][2^(2/3)x^2+2^(1/3)x(x-2)+(x-2)^2]=0 First term is real solution (SyberMath solution). Second term giver two complex conjugate solutions.

    • @SyberMath
      @SyberMath  3 роки тому +14

      I usually ignore the complex solutions but you're right!

  • @asimsinha2531
    @asimsinha2531 3 роки тому +3

    Beautiful way of solving the problem,enriching my knowledge.🙏🙏

  • @voltalimwabbit5492
    @voltalimwabbit5492 2 роки тому +3

    Solving this equation using lagrange’s resolvent and cardano’s formula in another video would be great sir!

  • @adityaghosh2844
    @adityaghosh2844 3 роки тому +5

    Sir, you have showed only the real solution and rationalised the denominator but there are also two other complex roots(also because by definition, a cubic equation has three roots) consisting of omega(cube root of 1), thank you for this solution....

    • @SyberMath
      @SyberMath  3 роки тому +2

      No problem. Yes there are complex solutions!

    • @alessandrofabi9252
      @alessandrofabi9252 3 роки тому +1

      Just to understand: what’s omega?

  • @JohnRandomness105
    @JohnRandomness105 2 роки тому +1

    Following my standard cubic method was less grungy than usual with this equation. I wound up with that answer, but I tried in vain to simplify it. I think that it took about as long as your method.

  • @tumak1
    @tumak1 3 роки тому +1

    ...nice job explaining the real number answer. Very smooth!

  • @moeberry8226
    @moeberry8226 3 роки тому +9

    Hands-down the best mathematician on UA-cam he and his problems are unmatched and every single problem applies so many different concepts. His explanation is always on point.

    • @scottleung9587
      @scottleung9587 Рік тому

      He's actually not a mathematician - he just enjoys solving math problems. However, he would be very flattered!

  • @garydudley6149
    @garydudley6149 Рік тому

    Magnificent mathematical innovation!

  • @mayankjangid1543
    @mayankjangid1543 3 роки тому

    A great one!

  • @eduardoteixeira869
    @eduardoteixeira869 3 роки тому +4

    Great solution. Thank you. I learned one more.

    • @SyberMath
      @SyberMath  3 роки тому +1

      Great to hear!

    • @leif1075
      @leif1075 2 роки тому

      @@SyberMath So far I got that this expression equals (x+2)^3 -12x but NOW WHAT..surely this is what most very intelligent people wkuld do..i havent watched your solution yet but what then do you have (x +2)^3= 24x and take the cube root of both sides?? What else i don't see why wohld multiply the equation by something because there's no pattern tnere or to dividie by anything..hope you can resoo d..

  • @DilipKumar-ns2kl
    @DilipKumar-ns2kl 3 роки тому

    You are amazing !

  • @damiennortier8942
    @damiennortier8942 2 роки тому

    Can you do more cubic equation with all type of solution please?

  • @franciscook5819
    @franciscook5819 Місяць тому

    x³+6x²-12x+8=0
    The coefficients hint at a perfect cube, dividing by the binomial coefficients ( 1 3 3 1)
    we get 1 2 -4 8 neatly powers of two (signs are a mess) then, as per video for 3 steps
    (x+2)³=x³+6x²+12x+8
    (x-2)³=x³-6x²+12x-8
    so 2x³-(x-2)³=x³+6x²-12x+8=0
    2x³=(x-2)³
    or ( (∛2.x)/(x-2) )³=1 (NB ∛ is the cube root - it doesn't look very readable on screen)
    Euler's formula: e^iθ = cosθ + i sinθ
    1=e^(2inπ) has (3rd) roots at n=0,1,2 i.e. e^0, e^(2iπ/3), e^(4iπ/3) so 1, -(1/2)±(i√3/2)
    for a solution s (one of the above)
    (∛2.x)/(x-2)=s
    x(∛2-s)=-2s
    x=2s/(s-∛2)
    s=1 => 2/(1-∛2) × (1+∛2+(∛2)²)/(1+∛2+(∛2)²)
    x=2(1+∛2+(∛2)²)/(1-2)
    x=-2-2^(4/3)-2^(5/3)
    s=-(1/2)±(i√3/2)
    x=2*(-(1/2)±(i√3/2))/(-(1/2)±(i√3/2)-∛2)
    =2*(-(1/2)±(i√3/2))/(-(1/2+∛2)±(i√3/2))
    after a whole mess of rearranging, the other solutions are
    x = -2+∛2(1+i√3)+(∛2)²(1-i√3)
    x = -2+∛2(1-i√3)+(∛2)²(1+i√3)

  • @fdh2277
    @fdh2277 2 роки тому

    Awesome!

  • @Drk950
    @Drk950 2 роки тому

    Nice trick!

  • @qwang3118
    @qwang3118 Рік тому

    One can divide the equation by 8 and let y = -x/2. So the equation becomes: -y^3 + 3y^2 + 3y + 1 = 0. 2y^3 = y^3 + 3y^2 + 3y + 1 = (y + 1)^3.
    2 = (1 + 1/y)^3. Taking cubic root, 1 + 1/y = r, where r = cubicRoot(2). There are 3 cubic roots of 2. r represents any of them.
    y = 1 / [r - 1] = [r^2 + r + 1 ]. As y = -x/2, x = -2 * [r^2 + r + 1 ].
    ...

  • @zidanihamid7485
    @zidanihamid7485 2 роки тому

    Beatifull amazing maths is beaty

  • @ananyagupta1409
    @ananyagupta1409 2 роки тому

    Great!

  • @christiansmakingmusic777
    @christiansmakingmusic777 Рік тому

    Very nice. If you don’t have a calculator handy you can approximate the root using recursion and the system y=24x, y=(x+2)^3 with x[0] less than zero.

    • @christiansmakingmusic777
      @christiansmakingmusic777 Рік тому

      More succinctly, just use the recursive relationship x[i+1]=-(1/24)(-x[i]+2)^3 with x[0]

  • @tbg-brawlstars
    @tbg-brawlstars Рік тому

    Nice!

  • @souzasilva5471
    @souzasilva5471 Рік тому

    Moço, você é maravilhoso. Conhece demais.(Brasil). (Young man, you are wonderful. You know too much.)

  • @vishalmishra3046
    @vishalmishra3046 3 роки тому

    How about this method of using (x+y)^3 pattern ?
    (x+2)^3 = x^3 + 3(x)(2)(x+2) + 2^3 = 24x = 24(x+2) - 48
    Let y = x+2 and m = -8 and n = -24. Then, y^3 + 3my = 2n.
    So, cubic-discriminant D = n^2 + m^3 = 576 - 512 = 64 = 8^2. Therefore x+2 = y = ∛(n + √D) + ∛(n - √D) = ∛(-24 + √64) + ∛(-24 - √64) = - ∛32 - ∛16
    Hence, *x = -2 - ∛32 - ∛16*

  • @wayneflanagin664
    @wayneflanagin664 2 роки тому

    All these different seemingly random equations with all these different tricks.I think this is crazy in one sense at least.Yes,algebraic skill is developed and that is laudable,but it makes me feel that something is amiss with all this stuff.Structural relations between different objects to find equation solving universality I crave.I pray to the mathematical gods to let me get to the bottom of all this.

  • @defeat7774
    @defeat7774 Рік тому

    5:45 This line seems like my native language dialogue 😂

  • @mohammadazadi4535
    @mohammadazadi4535 3 роки тому

    Especial thanks to thinking loudly for solving this. I like this method becsuse of being normal .thanks again.

  • @levitheentity4000
    @levitheentity4000 2 роки тому

    I lobe this sooooo much :D

  • @aniketchaudhary4013
    @aniketchaudhary4013 3 роки тому

    Thats great

  • @davidseed2939
    @davidseed2939 2 роки тому

    why the crossing out. the factor of (x-2) is clear. so (x-2)(x^2-4) +6x(x-2)=0

  • @imonkalyanbarua
    @imonkalyanbarua Рік тому

    So beautiful! 😇👏👏👏

  • @voltalimwabbit5492
    @voltalimwabbit5492 2 роки тому

    Could you solve this equation using cardano’s formula sir? Thank you a lot sir

  • @robyzr7421
    @robyzr7421 2 роки тому

    thx again dear teacher ..

  • @chimmychonga4795
    @chimmychonga4795 3 роки тому

    Alt.
    (2-x)^3+2x^3=0
    B/c you can flip the 2 to the front part of the equation. It's a bit more cleaner to.

    • @flanjunk
      @flanjunk 3 роки тому

      How would you know to arrange it that way?

    • @chimmychonga4795
      @chimmychonga4795 3 роки тому

      David Flanders Let 2=y, and notice how in a binomial expansion the signs alternate. Variables are just #'s.

  • @abesechinbasekar6252
    @abesechinbasekar6252 3 роки тому

    Can you suggest a book which has problems in algebra of this level

    • @SyberMath
      @SyberMath  3 роки тому +1

      Hard to find one book but I'm compiling a list of good books and will share them in the community tab soon. Feel free to remind me if I forget!
      ua-cam.com/channels/W4czokv40JYR-w7u6aXZ3g.htmlcommunity

  • @aliasgharheidaritabar9128
    @aliasgharheidaritabar9128 3 роки тому

    So neat.

  • @ak-indonesia4968
    @ak-indonesia4968 2 роки тому

    Unbelievable

  • @AnowarHossain-tz4hl
    @AnowarHossain-tz4hl 2 роки тому +1

    I am bangladeshi .but your class is very nice so I see and flow your lecture

    • @SyberMath
      @SyberMath  2 роки тому

      Many many thanks. Greetings from the United States! 💖

  • @sachinrajbhardseudwarka8246
    @sachinrajbhardseudwarka8246 3 роки тому

    smarter than smartest...
    i got solution
    thanks ❤️

    • @SyberMath
      @SyberMath  3 роки тому

      Thank you for the kind words! 💖

  • @kimba381
    @kimba381 4 місяці тому

    That's ONE or the solutions. What about the other 2?

  • @golakpatel1924
    @golakpatel1924 2 роки тому +1

    Simply multiply whole equation with (-1). Further the given equation must change in to (a-b) ^3 form

    • @leif1075
      @leif1075 2 роки тому

      Tjat doesn't do enough..that just gives you something that equals-(x -2)^3 but then what? I don't see what you can do

  • @MrLidless
    @MrLidless 2 роки тому

    Sub y = x + 2 to get y³ - 24y + 48 = 0 and Cardano it. I see the trick you were trying to show, but the messing around with the irrational denominator undid all the good work.

  • @thelogicone185
    @thelogicone185 2 роки тому

    Very good

  • @RamdevHub
    @RamdevHub 2 роки тому

    -2 is also a real solution how to find that

  • @antoine5571
    @antoine5571 3 роки тому

    awesome

    • @SyberMath
      @SyberMath  3 роки тому +1

      Thank you, Antoine! 🤩

  • @sebastianviacava743
    @sebastianviacava743 2 роки тому

    You hsce x minus a multiply x minus b multiply x minus c is equal to that cubic equstiion too

  • @user-ph4zh1lm8w
    @user-ph4zh1lm8w Місяць тому

    You are best

  • @geekchick4859
    @geekchick4859 2 роки тому

    Why didn’t you just find one factor then use long division?? Would have been quicker than all this mucking about.

  • @hamidrezax34
    @hamidrezax34 3 роки тому

    -7.69

  • @barakathaider6333
    @barakathaider6333 2 роки тому

    👍

  • @saikrishnamohanrao5246
    @saikrishnamohanrao5246 3 роки тому

    Abrakadabra....🤣

    • @SyberMath
      @SyberMath  3 роки тому

      😂 Ehehe! This is like magic, isn't it?

  • @rocamgreg
    @rocamgreg 2 роки тому

    👍👍👍

  • @calvinjackson8110
    @calvinjackson8110 2 роки тому

    Tricks work if somebody gives you a cubic where they work. Chang one coefficient or a sign and you may have to throw in the towel. Give me something that works ALL the time, without tricks.

  • @geekchick4859
    @geekchick4859 2 роки тому

    Please don’t write at the bottom of the screen. We can’t see under the subtitles.

    • @SyberMath
      @SyberMath  2 роки тому

      I keep forgetting 😁

  • @ashutoshmeena2311
    @ashutoshmeena2311 2 роки тому

    it was lengthy problem and not tricky

  • @kfjfkeofitorhf9520
    @kfjfkeofitorhf9520 Рік тому

    3X+12X-12X+8=0

  • @tajbibishamim8085
    @tajbibishamim8085 2 роки тому

    In exam you do not have so much time to try all that

    • @SyberMath
      @SyberMath  2 роки тому

      Right but you have time if you're not taking a test

  • @nadiamaliq6502
    @nadiamaliq6502 Рік тому

    apenii😂xpahamla

  • @vivekbhutada3049
    @vivekbhutada3049 3 роки тому +1

    Mathemagic👍

  • @ferdithus
    @ferdithus 2 роки тому

    This is wrong, x should have 3 answers. Not one answer. x=2, x=2(sqrt(3)-2) , x=-2(sqrt(3)+2)

  • @carloshuertas4734
    @carloshuertas4734 3 роки тому

    I got x=-2.

    • @SyberMath
      @SyberMath  3 роки тому

      How?

    • @carloshuertas4734
      @carloshuertas4734 3 роки тому

      -(2*2*2)+6*-1(2*2)-12(-2)+8=0. That's how I got x=-2. Check it out.

    • @carloshuertas4734
      @carloshuertas4734 3 роки тому

      It's -8-24+24+8=0

    • @Salarr
      @Salarr 3 роки тому

      @@carloshuertas4734 doesn't work

    • @Salarr
      @Salarr 3 роки тому

      @@carloshuertas4734 you multiplied wrong on the x² term

  • @daddykhalil909
    @daddykhalil909 3 роки тому

    For God’s sake remove the subtitles especially that you are talking English!

    • @SyberMath
      @SyberMath  3 роки тому +1

      You can turn off the subtitles