A Fun Way to Solve Cubics: Vieta's Substitution

We explore an interesting method of solving cubic equations, using a clever substitution. The method presented works for any cubic, but may require the extra step of writing in the form x^3 + ax + b = 0. See here:

00:00 Intro
00:20 Example
04:15 Solutions
07:43 The general method
11:01 Why don't we get 6 solutions?

[safekid]

Your channel is the best thing to happen to maths since the Leibniz rule of integration

[asklar]

I'm fascinated by Viète's story, especially about how little credit he seems to get considering his monumental achievements; but all those pale in comparison to the stories of intrigue, code-cracking/cryptography, and murder. I need a Netflix show about Viète's life!

[spaghettiking]

Thanks, this video is lovely! The only method I knew of to solve cubics was just guess+check using the rational roots theorem, which also doesn't work if the roots aren't rational... very helpful!

[vvop]

I was mesmerised throughout. Brilliant, well done, it's now my new favourite channel!

[heartache]

i've never heard of this method before, it looks so simple

[陈黎炜]

I met this problem when reading Visual Complex Analysis, then I went to Youtube, then i met you!

[rubensramos]

Nice. Well done. The real root of x^3+6x-2=0 is simply x =sqrt(3Wq(1/27)) = 0.3275, where Wq is the Lambert-Tsallis function with q = 1/2 in this case. One can find more about the Lambert-Tsallis function in this minipaper: "Solving the Fractional Polynomial a(x^r)+b(x^s)+c = 0 Using the Lambert-Tsallis Wq Function" that one can find on Researchgate.

[jolliet]

Great video! You've solved my problem in a very elegant way! Love the w notation!

[ayliose]

Thank you very much for this explanation! Trying to code a cubic solver without using premade packages.

[bjornfeuerbacher]

Or more generally, for the equation ax³ + bx² + cx + d = 0, use the "simple" substitution x = z + (b³-3ac)/(9a²z) - b/(3a). ;)

[dylwhs]

This is the first time I have heard of imaginary roots of unity in relation to cubic roots. Thanks! Its also the first time I have seen Vieta's substitution explained. 😃👍🏻

[holyshit]

I prefer Fontana's substitution x=u+v (I'm not sure del Ferro solved the same way)
There is nice geometrical interpretation with volumes
Euler generalized Fontana's method to quartic
I do not see how Vieta subsittution can be generalized to quartic

[xactxx]

It's nice to see here the real reason why complex numbers, historically, started to be taken seriously - they are essential in the solution of cubic equations.

[holyshit]

I inspired by the substitution found in Fichtenholz book Course on differential and integral calculus (Курс дифференциального и интегрального исчисления in original)
in the paragraph about elliptic integral tried to derive my own method for quartic
Idea behind this method was that i tried to reduce general quartic to the biquadratic which can be reduced to quadratic by simple substitution
In quartic
i substituted x = (pt+q)/(t+1) where p and q are undetermined coefficients
After this substitution I equated coefficients in terms t^3 and t to zero
and got system of equation
Solution of this system of equations lead me to the polynomial equation of 10th degree but polynomial
was a divisor of this 10th degree polynomial so i left with sextic which is difficult to solve for me

[Amoeby]

You could use that cbrt(t_1) * cbrt(t_2) = -2 where t_1 and t_2 are solutions to t^2 - 2t - 8 = 0 and t = z^3 to find the right pairs t_1 and t_2. Final solutions would be in the t_1 + t_2 form.

[joningram]

Thank you for the video - very nicely explained. I wonder what the motivation for the substitution was. It seems like a real stroke of luck that the same value of k can be used to remove two terms at the same time.

I also like the colour scheme!

[adandap]

Very nicely explained - thank you. The ending was a tad abrupt though, so maybe you should say something like "and that's a good place... for a cup of tea" or something. 🙂

[bingeu]

A good method indeed. But are there any faster method to solve the cubic equations?

[davidbrisbane]

Cardano's depressed (real) cube root formula tells is that if x³ + px + q = 0, then x is a real root of this equation
where x = u + v and p and q are real, such that
u = ∛[-q/2 + √[(q/2)² + (p/3)³]] and
v = ∛[-q/2 - √[(q/2)² + (p/3)³]]

The above is the usually stated form of the real cube root, however, during the derivation of this root x = u + v, we discover that v = -p/3u.

Hence x = u + v can be written as
x = u + (-p/3)*(1/u)
Now if we let z = u
⇒ x = z + (-p/3)*(1/z), which is Viéte's substitution,
which in the case x³ + 6x - 2= 0, has a substitution,
x = z - 2/z.

[actuarialscience]

I hate trivial and "trial and error" mathematics. That is why I love this video.