A Fun Way to Solve Cubics: Vieta's Substitution

[@safekid01]

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

[@DrBarker]

Haha, thank you!

[@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!

[@davidbrisbane7206]

Viète also had a big influence on the beginning of standard notation for variables in equations before Descarte developed (more or less) our current symbolic approach to writing equations in analytic geometry. Also, Vèite had a big influence on Fermat's work in analytical geometry.

[@davidbrisbane7206]

Indeed he did.

[@spaghettiking653]

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!

[@davidbrisbane7206]

We can use *Descartes' Rule of Signs* to determine how many positive real roots and how many negative real roots a polynomial has. Applied to x³ + 6x - 2 = 0, then there is one positive real root, and zero negative roots. It's clear that if f(x) = x³ + 6x - 2, then f(1/6) < 0 and f(1/3) > 0. So by the Intermediate Value Theorem (IVR), there exists a point c, such that f(c) = 0. So c is the root of f(x). So, approximating the root of f(x) to be x₀. Ie, f(x₀) = 0 (approximately) and then use the *Newton-Raphson method* to find x₁, x₂, x₃ etc, which eventually are ever closer to the exact value of the root, ie c, then we can find this root to any number of decimal places as desired.

[@vvop]

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

[@DrBarker]

Thank you!

[@heartache5742]

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

[@davidbrisbane7206]

Looks simple, but it can be a horrific exercise to simplify the roots.

[@heartache5742]

@@davidbrisbane7206 counterpoint, it's the cubic, it can't be too easy

[@davidbrisbane7206]

@@heartache5742 Suppose x³ - 7x + 6 = 0. Almost by inspection, the roots are x = 1, 2 and -3. Now try solving this equation using Viéte's substitution (or Cardano's method). Not trivial to recover these three integer roots using these methods.

[@heartache5742]

@@davidbrisbane7206 that's why you always check 1

[@陈黎炜]

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

[@jolliet1009]

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

[@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. 😃👍🏻

[@experimentingalgorithm1546]

You can read a little book called higher Algebra By Hall and Knight to get more acquainted with it

[@bingeu5819]

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

[@DrBarker]

Yes, there is a general cubic formula, or we could use Cardano's formula for a cubic once we write it in the form x^3 + ax + b = 0. This method isn't the fastest, but I like how it gives a sense of having "earned" the solution by doing the calculations. Kind of like solving a quadratic by completing the square and then solving algebraically, vs. just using the quadratic formula.

[@ayliose9188]

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

[@DrBarker]

Sounds like a fun project!

[@rubensramos6458]

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.

[@uzumakiboruto7194]

Love your channel man❤

[@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.

[@ronecisilva4143]

Parabéns pela explicação. Fascinante!

[@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!

[@bjornfeuerbacher5514]

One can arrive at that substitution by studying how Tartaglia and others solved cubic equations. Look up "Cardano's equation" and how it was found. Or for a short summary, you could look at David Brisbane's comment below.

[@qwertyuiop2161]

so cool, earned yourself a sub

[@scottleung9587]

Cool - this is a great method if the rational root theorem or factoring aren’t easy to use. BTW what does the general solution look like for quartics and higher degree polynomials?

[@michaeltuchman9656]

There is no general solution to quintics and higher-degree polynomials. This is a consequence of the Abel-Ruffini theorem, which you should google. Notice that in the video's problem, there was a 6th-degree polynomial, but could be solved like a quadratic in z^3. For the 4th degree, you get a 24th-degree polynomial that can be solved as a cubic. However, for the 5th degree, you wind up with a 120th-degree polynomial that can't be regrouped as a lower-degree polynomial in z^5. You're stuck. It's more complicated than that, but that's where you should start. Enjoy.

[@holyshit922]

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

[@DrBarker]

That's another really nice method! I don't think Vieta's substitution can be generalised, but it would be interesting to see if there is something similar for higher order equations.

[@holyshit922]

@@DrBarker Let's look at Eulers generalisation of Fontana's method We assume that root of quartic is a sum of three terms x = u+v+w and then rewrite equation as system of equation which can be easily transformed into Vieta formulas for cubic In fact resolvent will be sextic with non zero terms only for even powers In this system which i mentioned there is product uvw = something and maybe from this we could derive substitution similar to Vieta substitution for cubics

[@marcbollee8052]

Great video, thank you!

[@bjornfeuerbacher5514]

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

[@scottleung9587]

Thanks so much!

[@XJWill1]

Not quite. It should be b^2 , not b^3

[@bjornfeuerbacher5514]

@@XJWill1 Thanks for correcting my typo. :)

[@77Chester77]

Nice method and good presentation

[@edmundwoolliams1240]

Is this where the cubic formula comes from?

[@DrBarker]

This substitution method gives us a way of deriving/understanding the cubic formula, but there are other ways to derive the formula too.

[@actuarialscience2283]

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

[@holyshit922]

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 a_{4}x^4+a_{3}x^3+a_{2}x^2+a_{1}x+a_{0}=0 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 a_{4}p^4+a_{3}p^3+a_{2}p^2+a_{1}p+a_{0} was a divisor of this 10th degree polynomial so i left with sextic which is difficult to solve for me

[@DrBarker]

I guess for a quartic, there might be too many coefficients for us to be able to find a substitution that turns it into something significantly simpler to solve? Maybe this sort of approach will work though for special cases of quartics analogous to the special case of depressed cubics (x^3 + ax + b = 0)?

[@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.

[@yt-1161]

What is your field of interest Dr. ? Number theory

[@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. 🙂

[@DrBarker]

Haha, I like this suggestion!

[@昆仑云路]

A good new method!

[@davidbrisbane7206]

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.

[@picrust314]

Good presentation. But how would you solve x^3-7x+6=0 with a simular substitution? You’ll end up with expressions that are not easily shown to be the acctual integer solutions.

[@davidbrisbane7206]

First, note that using the Rational Roots theorem, we can show that if rational roots exist, then they must be x = ±1, ±2, ±3, or ± 6. If we try each of these possibilities in x³ - 7x + 6 = 0, we find that the solutions are x = -3, 1, and 2. Thus we have found three roots and that is all the roots, as a cubic equation has three roots. Now let's try and solve x³ - 7x + 6 = 0 using Viéte's substitution method. The appropriate substitution is x = z + 7/3z. Now x³ - 7x + 6 = 0 and x = z + 7/3z ⇒ x³ - 7x + 6 = (z + 7/3z)³ - 7(z + 7/3z) + 6 = 0 ⇒ z³ + 343/27z³ + 6 = 0 (after simplification) ⇒ 27z⁶ + 162z³ + 343 = 0 ⇒ 27(z³)² + 162z³ + 343 = 0 ⇒ z³ = [-162 ± √(162² - 4(27)(343))]/[2(27)] ⇒ z³ = [-162 ± √(-10,800)]/54 ⇒ z³ = [-162 ± i√(10,800)]/54 ⇒ z³ = [-162 + i√(10,800)]/54, by just taking the positive root, as the negative root results in the same cube roots for z. ⇒ z = z₁ = ∛[[-162 + i√(10,800)]/54] ⇒ z₁ = ∛[[-81 + i√2700]/27] Now ∛[[-81 + i√2700]/27] = ∛[[-81 + 30i√3]/27] Now assume [-81 + 30i√3] = (a + bi√3)³ , where a, b ∈ ℤ. Clearly a = 0, or b = 0, has no solution. This is a critical assumption. We know by the form of a root (say x₁) that we have a good chance that when z₁ and 7/3z₁ are added together that the imaginary parts will cancel each other. So [-81 + 30i√3] = (a + bi√3)³ = (a³ - 9ab²) + (3a² - 3b³)√3 after simplification ⇒ -81 = a³ - 9ab² & 10 = b(a² - b²) Now 10 = b(a² - b²) ⇒ b = ±1, ±2, ±5 or ±10, as b divides 10. After going through the above possibilities for b, we find that the only possibility is b = 2, which requires that a = 3. Now a³ - 9ab² and a = 3 and b = 2 ⇒ a³ - 9ab² = 3³ - 9(3)(2²) = -81. So indeed, a = 3 and b = 2 is a solution to -81 = a³ - 9ab² & 10 = b(a² - b²) ⇒ [-81 + 30i√3]/27= [(3 + 2i√3)/3]³ ⇒ ∛[[-81 + 30i√3]/27] = (3 + 2i√3)/3 ⇒ z₁ = (3 + 2i√3)/3 Now 7/3z₁ = (7/3)[3/(3 + 2i√3)] = (3 - 2i√3)/3, after simplification Hence x₁ = z₁ + 7/3z₁ = (3 + 2i√3)/3 + (3 - 2i√3)/3 = 2 Let's find the other two cube roots x₂ and x₃. Let 1, ɷ, and ɷ² be the cube roots of 1. We can compute ɷ = (-1 + √3)/2 and ɷ² = (-1 - √3)/2 and 1/ɷ = ɷ² and 1/ɷ² = ɷ. Hence, x₂ = z₁ɷ + 7/(3z₁ɷ) = z₁ɷ + [7/(3z₁)]ɷ² = -3 x₃ = z₁ɷ² + 7/(3z₁ɷ²) = z₁ɷ² + [7/(3z₁)]ɷ = 1 So, x₁ = 2, x₂ = -3 and x₃ = 1. As you can see, it's not obvious how to recover the integer solutions, but it is possible. If Candano's method had been used to find the cube roots of the depressed cubic, then we'd follow a similar, but not identical approach for finding the roots of the depressed cubic.

[@picrust314]

@@davidbrisbane7206 Thank you for showing the last steps. Not too obvious how to get to the nice integer solutions.

[@davidbrisbane7206]

@@picrust314 👍

[@RexxSchneider]

@@davidbrisbane7206 Wouldn't Vieta's substitution suggest the substitution x = z + 7/3z rather than x = z +7/z?

[@davidbrisbane7206]

@@RexxSchneider Look @9:40 where k = -a/3

[@antoniomontana2020]

You are incredible

[@kaizoisevil]

See, for cubics without the x^2 term, I thought you could try answers in the form of cbrt(r)+cbrt(s)

[@prldh]

If you substitute x=z-p/(3z) it works for every third grade depressed equation x^3+p*x+q=0 Note that in this case p/3 = 6/3 = 2

[@sneakylemon8513]

Thank you

[@Deepak0Aggarwal]

Why x is z -2/z

[@DrBarker]

The idea is that if we choose our z-substitution carefully enough, we can turn our cubic equation into a quadratic equation in the variable z^3, which is then easier to solve. For other cubic equations, we can use the formula from the second half of the video.

[@bosorot]

x=z-b/3z . this is a general formula . b is a number in front of x . In this case is 6 . so x=z-6/3z or z-2/z .

[@raystinger6261]

So basically we're increasing the multiplicity of each root by one with this substitution. I'm here wondering how we could expand on that.

[@AitJoseph]

You say solve ! in R or C ?. Wath's the problem ?

[@LifeIsBeautiful-ki9ky]

This Method was Alredy discovered by Francis Viete.

[@markregev1651]

Do UK unis use w instead of omega?

[@RexxSchneider]

Not necessarily. The Greek letter ω is commonly used for the principal cube root of unity, but it actually doesn't matter what letter you use to stand for e^(2πi)/3, does it?

[@leesweets4110]

Cardanos substitution, yo. Vieta's contribution to this was trivial. And Vieta actually does have his own solution to the cubic; this is not it.

[@MultiChrisjb]

Oh, it's just too much fun for me. I'd stop at all real solutions.

[@nicolastorres147]

The last minute means we can always ignore one of the two z values

[@lindermartinezgalvez2444]

Interesante 😊, lo resolví en la universidad un problema parecido, me hizo recordar la época universitaria.

[@Dharmarajan-ct5ld]

Should you not give logic behind the particular substitution ...The form, constants and validity. It would then be educative. Please consider

[@RexxSchneider]

The entire second half of the video is doing precisely that.

[@Guimaster127]

I may be missing something here, but why do we know that alpha^3*beta^3 = -a^3/27?

[@bjornfeuerbacher5514]

Use Vieta's formula for quadratic equations: The equation x² + px + q = 0 having the solutions x1, x2 is equivalent to q = x1 x2 (and -p = x1 + x2).

[@ved9402]

It would be better to spell it as omega

[@RexxSchneider]

Not really. You'd just get as many comments asking why he was pronouncing 'w' as omega.

[@vishalmishra3046]

*Simply use the General Cubic Formula* IF { x^3 + 3 m x = 2 n } THEN { x = (n + d)^(1/3) + (n - d)^(1/3) where d^2 = n^2 + m^3 } Here, x^3 + 3 (m=2) x = 2 (n=1), therefore d^2 = n^2 + m^3 = 1^2 + 2^3 = 1 + 8 = 9 = 3^2, therefore n+d = 1+3 = 4 and n-d = 1 - 3 = -2 Therefore, x = (n + d)^(1/3) + (n - d)^(1/3) = 4^(1/3) - 2^(1/3). This is the only real solution. The remaining 2 will be complex conjugate solutions using cube-roots of unity, w = e^(i 2π / 3) and w' = it's complex conjugate, (1 ± i√3)/2. Hence, the 3 solutions are, x = (A - B, Aw - Bw', Aw' - Bw) where, A^3=4 and B^3=2 and A,B are both real. Substituting back into the equation confirms that all 3 solutions are distinct and correct.

[@knightofcarrion7358]

What happens when you over complicate things:

[@gemerson7406]

Os gringos são melhores que os brasileiros! (Bom, a maioria sim)

[@holyshit922]

This substitution will not work for cubics x^3+b = 0 Moreover you have to be careful because of possible division by zero

[@DrBarker]

This is a good point. Fortunately, a cubic of the form x^3 + b = 0 is nice and easy to solve.

[@holyshit922]

Problem for you Prove that there is no general way solving quartic equation which avoids solution of cubic equation so called cubic resolvent

[@DeMoNiC_pEpPa_PiG]

Synthetic division is easier…