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

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

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

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.