Solving cubic polynomials

There are many sources for how to solve cubic polynomials, often with many peripheral details, so a summarised and simplfied version is helpful when implementing a new code. This page is an attempt to record that.

General solution

We will consider the monic cubic polynomial in <math>x</math>, with <math>a</math>, <math>b</math> and <math>c</math> real-valued:

<math> x^3 + a x^2 + b x + c = 0 </math>

To solve for <math>x</math>, we first calculate the coefficients of an equivalent 'depressed cubic' <math>y^3 - 3P y - 2Q = 0</math>, by substituting <math>x = y - a/3</math>:

<math>3P = \frac{a^2}{3} - b</math>

<math>2Q = \frac{2}{27} a^3 - \frac{1}{3} ab + c</math>

If <math>P</math> and <math>Q</math> are both zero, then we can stop here with a triple root <math>y = 0</math>, that is, <math>x = -\frac{a}{3}</math>.

If only <math>P</math> is zero, then <math>y^3 = 2Q</math>, so <math>x = \sqrt[3]{2Q}-\frac{a}{3}</math>.

If only <math>Q</math> is zero, the depressed cubic is simply <math>y^3 - 3Py = y \left(y^2 - 3P \right) = 0</math>, which has a real root at <math>y=0</math> and two more at <math>y = \pm \sqrt{3P}</math>, which may be either real or imaginary, depending on the sign of <math>P</math>.

Otherwise, the depressed cubic can then be converted into a quadratic in <math>z^3</math> via the 'magic' Vieta substitution <math>y = z - \frac{P}{z}</math>,

<math>\left(z^3\right)^2 - 2Q \left(z^3\right) + P^3 = 0</math>

for which the solution is

<math>z^3 = Q \pm \sqrt{Q^2 - P^3}</math>

(1)

The original cubics (in <math>x</math> and <math>y</math>) have real coefficients, which means that any complex roots must appear with their complex conjugate. As such, the cubic equations must have either one or three real roots, and two or zero complex roots. The determinant <math>\Delta = Q^2 - P^3</math> is critical in determining which case arises.

Case of negative determinant <math>\Delta<0</math>

If <math>\Delta < 0</math> then <math>P^3</math> and <math>P</math> must also be negative, and we must solve

<math>z^3 = R</math> or <math>z^3 = \overline{R}</math>

where

<math>R = Q + i \sqrt{-Q^2 - P^3}</math>

<math>\left|R\right| = \sqrt{Q^2 - Q^2 - P^3} = \sqrt{-P^3} </math>

<math>\arg R = \theta = \cos^{-1} \left(\frac{Q}{\sqrt{-P^3}}\right)</math>

and the solution for <math>z</math> is then

<math>z = \sqrt{-P} \cos\left(\frac{\theta + 2\pi k}{3}\right) + i \sin\left(\frac{\theta + 2\pi k}{3}\right)</math> where <math>k \in \left\lbrace 0,1,2 \right\rbrace</math>, for <math>z^3 = R</math>

or

<math>z = \sqrt{-P} \cos\left(\frac{-\theta + 2\pi k}{3}\right) + i \sin\left(\frac{-\theta + 2\pi k}{3}\right)</math> where <math>k \in \left\lbrace 0,1,2 \right\rbrace</math>, for <math>z^3 = \overline{R}</math>

Clearly the these solutions are conjugates of each other.

Evaluating <math>y</math>, we have

<math> y =

[3]{\left|R\right|}~\mathrm{cis}\left(\frac{\theta}{3} + \frac{2\pi}{3}k\right)</math> where <math>k \in \left\lbrace 0,1,2 \right\rbrace</math>

, and determines the nature of the roots of the equation.

If <math>\Delta</math> is negative, the right hand size of (1) is a complex conjugate pair. Taking the cube root results in six different values (three from each conjugate). When those six values are used to calculate <math>y</math>, the result is:

In the general case, the right side of this equation <math>R = Q \pm \sqrt{\Delta}</math> is complex, and can be written as

<math>z^3 = r~\mathrm{cis}~\theta = r \cos \theta + i r \sin \theta</math>

where <math>r = \left|R\right|</math>, and <math>\theta = \arg R</math>.

<math> \sqrt[3]{z} = \sqrt[3]{r} \mathrm{cis}~\left(\frac{\theta}{n} + \frac{2\pi}{n} k \right)</math> for <math>k \in \left\lbrace 1,2,...,n-1 \right\rbrace</math>

If <math>\Delta = 0</math>, then our solution is the real equation

<math>z^3 = Q</math>

which, propagating back to the original cubic, gives the triple root

<math>x = -\frac{a}{3} + \sqrt[3]{Q} - \frac{P}{\sqrt[3]{Q}}</math>

Calculation checking

The following sympy code can be used to confirm the above results:

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