Solving a Cubic equation Using an Algebraic Trick

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[nilsastrup]

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.

[damiennortier]

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]

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.

[misterdubity]

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

[mathsfamily]

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

[esteger]

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.

[asimsinha]

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

[golddddus]

8:18 There are two more solutions. First term is real solution (SyberMath solution). Second term giver two complex conjugate solutions.

[tumak]

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

[voltalimwabbit]

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

[JohnRandomness]

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.

[eduardoteixeira]

Great solution. Thank you. I learned one more.

[adityaghosh]

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....

[moeberry]

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

[defeat]

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

[christiansmakingmusic]

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.

[damiennortier]

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

[qwang]

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 ].
...

[souzasilva]

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

[franciscook]

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)


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)