# How to solve a cubic equation

Consider the general cubic equation

$x^3+bx^2+cx+d=0.\quad\quad\quad(*)$

Let

$y = x + \omega \implies x = y - \omega.\quad\quad\quad(1)$

Substituting (1) into (*), we have

$(y-\omega)^3 + b(y-\omega)^2+c(y-\omega) + d = 0.$

This is a new cubic equation in $y$:

$y^3 + (b-3\omega)y^2 + (3\omega^2-2b\omega+c)y - \omega^3+b\omega^2-c\omega+d=0\quad\quad\quad(2)$

Let

$\omega = \frac{b}{3}\quad\quad\quad(3)$

so that (2) becomes

$y^3 + (c-\frac{b^2}{3})y+d-\frac{bc}{3}+\frac{2b^3}{27}=0.$

That is

$y^3+py=q\quad\quad\quad(4)$

where $p = c-\frac{b^2}{3}, q = -d+\frac{bc}{3}-\frac{2b^3}{27}$ (see Fig. 1).

Thus, (4) is the so called depressed cubic equation. A solution is readily available through Cardano’s formula (see “Through the Mind’s Eye“).

Once we know $y$, (1) gives

$x = y - \frac{b}{3}.$

Fig. 1

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