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

x^3 + cx + d = 0

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.


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

27 a^2 c+81 a d-9 c^2 = 0

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.


[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.