# The Cubic Formula Derivation

Consider the arbitrary cubic equation $ax^3 + bx^2 + cx + d = 0$ for real numbers $a$, $b$, $c$, $d$ with $a\neq0$. By the fundamental theorem of algebra this equation has three roots $x_1$, $x_2$, $x_3$, over the complex numbers. Using the factor theorem gives the factorization $ax^3 + bx^2 + cx + d = a(x-x_1)(x-x_2)(x-x_3) .$ Expanding out the right-hand side gives $ax^3 - a(x_1+x_2+x_3)x^2 + a(x_1x_2+x_1x_3+x_2x_3)x - ax_1x_2x_3 ,$ and equating coefficients with the original expression gives the following system of equations: \begin{align*} b &= - a(x_1+x_2+x_3) \\ c &= a(x_1x_2+x_1x_3+x_2x_3) \\ d &= - ax_1x_2x_3 \end{align*} And now it is non-obvious how to proceed and solve for $x_1$, $x_2$, $x_3$.

## Introduction of New Variables

It turns out that it is fruitful to introduce the new variables $y_1$, $y_2$ by the definitions \begin{align*} y_1 &= a(x_1+\omega x_2+\omega^2 x_3) \\ y_2 &= a(x_1+\omega^2 x_2+\omega x_3) \end{align*} where $\omega=e^{2\pi i/3}=(-1+\sqrt{-3})/2$ is a cube root of unity. Then the following identities are satisfied: \begin{align*} y_1^3+y_2^3 &= -2b^3 + 9abc - 27a^2d \\ y_1y_2 &= b^2 - 3ac \end{align*} These can be found using Maple, for example.

## Solving for the Variables

Note that $y_1^3$, $y_2^3$ are the roots of the quadratic polynomial $(y-y_1^3)(y-y_2^3)=y^2-(y_1^3+y_2^3)y+(y_1y_2)^3$. In light of the above identities, this polynomial can be written as $y^2 - (-2b^3+9abc-27a^2d)y + (b^2-3ac)^3 .$ From which the roots $y_1^3$ and $y_2^3$ can be found using the quadratic formula. Taking cube roots gives $y_1$ and $y_2$, although only 3 of the 9 possible choices for cube roots will work correctly; namely those which satisfy the identity $y_1y_2 = b^2 - 3ac$. This identity will not always be satisfied if one simply uses principal roots, but one can use a modified principal root function.

Once $y_1$ and $y_2$ have been determined, one can solve the linear system \begin{align*} b &= - a(x_1+x_2+x_3) \\ y_1 &= a(x_1+\omega x_2+\omega^2 x_3) \\ y_2 &= a(x_1+\omega^2 x_2+\omega x_3) \end{align*} for $x_1$, $x_2$, and $x_3$. The solution is given by \begin{align*} x_1 &= (-b+y_1+y_2)/(3a) \\ x_2 &= (-b+\omega^2 y_1+\omega y_2)/(3a) \\ x_3 &= (-b+\omega y_1+\omega^2 y_2)/(3a) \end{align*} or more compactly by $x = \frac{-b+\omega^n y_1+\omega^{2n} y_2}{3a}$ for $n=0$, $1$, $2$.