How to solve a cubic equation

The process for solving a cubic equation seems like a sequence of mysterious tricks. I’d like to try to make the steps seem a little less mysterious.

Depressed cubic

The previous post showed how to reduce a general cubic equation to one in the form

which is called a “depressed cubic.” In a nutshell, you divide by the leading coefficient then do a simple change of variables that removes the quadratic term.

Now what? This post will give a motivated but completely ahistorical approach for removing the linear term cx.

Resultants

Suppose we don’t know how to solve cubic equations. What do we know how to solve? Quadratic equations. So a natural question to ask is how we might find a quadratic equation that has the same roots as our cubic equation. Well, how can you tell in general whether two polynomials have a common root? Resultants.

This is the point where we completely violate historical order. Tartaglia discovered a general solution to depressed cubic equations in the 16th century [1], but Sylvester introduced the resultant in the 19th century. Resultants were a great idea, but not a rabbit out of a hat. It’s not far fetched that some sort of determinant could tell you whether two polynomials have a common factor since this is analogous to two sets of vectors having overlapping spans. I found the idea of using resultants in this context in [2].

Tschirnhaus transformation

In 1683, Tschirnhaus published the transform that in modern terminology amounts to finding a polynomial T(x, y) that has zero resultant with a depressed cubic.

Tschirnhaus assumed his polynomial T has the form

$T(x, y) = x^2 + a x + \frac{2}{3} c + y$

Let’s take the resultant of our cubic and Tschirnhaus’ quadratic using Mathematica.

    Resultant[x^3 + c x + d, x^2 + a x + 2 c/3 + y, x]

This gives us

$y^3 + \frac{1}{27} \left(27 a^2 c+81 a d-9 c^2\right)y + \frac{1}{27} \left(-27 a^3 d+18 a^2 c^2+27 a c d+2 c^3+27 d^2\right)$

which is a cubic equation in y. If the coefficient of y were zero, then we could solve the cubic equation for y by simply taking a cube root. But we can make that happen by our choice of a, i.e. we pick a to solve the quadratic equation

So we solve this equation for a, plug either root for a into the expression for the resultant, then solve for y. Then we take that value of y and find where Tschirnhaus’ polynomial is zero by solving the quadratic equation

$x^2 + a x + \frac{2}{3} c + y = 0$

We solved for a value of y that makes the resultant zero, so our original polynomial and Tschirnhaus’ polynomial have a common root. So one of the roots of the equation above is a root of our original cubic equation.

Footnotes

[1] In this blog post, we first reduced the general quadratic to the depressed form, then solved the depressed form. This isn’t the historical order. Tartaglia came up with a general solution to the depressed cubic equation, but was not able to solve equations containing a quadratic term.

[2] Victor Adamchik and David Jeffrey. Polynomial Transformations of Tschirnhaus, Bring and Jerrard. ACM SIGSAM Bulletin, Vol 37, No. 3, September 2003.