Cubic Equation

• Case 1. $p = 0$ or $q = 0$
• Case 2. $\Delta = 0$ (All roots are real and two equal)
• Case 3. $\Delta > 0$ (One is a real root and two complex roots)
• Case 4. $\Delta < 0$ (All roots are real and different)
  • Alternative Form
• References

The classic technique to solve a cubic equation $$\begin{aligned} x^3 + ax^2 + bx + c = 0 \end{aligned}$$

is called Cardano’s formula, which reduces the original equation to the more simple form. $$\begin{aligned} t^3 + pt + q = 0 \end{aligned}$$

Once the solution $t$ is found, $x$ can be naturally obtained from $x = t - a/3$. So, replacing $x$ by $t - a/3$ in the original equation to find $p$ and $q$, $$\begin{aligned} &\left( t - \frac{a}{3} \right)^3 + a \left( t - \frac{a}{3} \right)^2 + b \left( t - \frac{a}{3} \right) + c \\\\ = &\left( t^3 - 3t^2 \frac{a}{3} + 3t \frac{a^2}{9} - \frac{a^3}{27} \right) + a \left( t^2 - 2t \frac{a}{3} + \frac{a^2}{9} \right) + b \left( t - \frac{a}{3} \right) + c \\\\ = &\; t^3 + \left( b - \frac{a^2}{3} \right)t + \frac{2a^3}{27} - \frac{ab}{3} + c = 0 \\\\ \Longrightarrow &\quad p = b - \frac{a^2}{3} \qquad q = \frac{2a^3}{27} - \frac{ab}{3} + c \end{aligned}$$

Note that this note is only interested in finding real roots.

Case 1. $p = 0$ or $q = 0$

• If $p = q = 0$, the reduced form becomes $t^3 = 0$, which produces $x = -a/3$.

• If $p = 0$ and $q \ne 0$, the reduced form becomes $t^3 + q = 0$, which produces $x = \sqrt[3]{-q} - a/3$.

• If $q = 0$ and $p \ne 0$, the reduced form becomes $t^3 + pt = t(t^2 + p) = 0$, which produces $x = -a/3$, or $x = \pm \sqrt{-p} - a/3$ if only $p < 0$.

In the case of $p \ne 0$ and $q \ne 0$, the following identity can be considered. $$\begin{aligned} (u - v)^3 + 3uv(u - v) = u^3 - v^3 \end{aligned}$$

This identity can be reformed to $t^3 + pt + q = 0$ form by setting $t = u - v$, $p = 3uv$, and $q = v^3 - u^3$. Since $p \ne 0$ and $q \ne 0$, $$\begin{aligned} v = \frac{p}{3u} &\Longrightarrow q = v^3 - u^3 = \left( \frac{p}{3u} \right)^3 - u^3 \Longrightarrow u^3 + q - \left( \frac{p}{3u} \right)^3 = 0 \\\\ &\Longrightarrow u^6 + qu^3 - \left( \frac{p}{3} \right)^3 = 0 \\\\ &\Longrightarrow u^3 = \frac{-q \pm \sqrt{q^2 + 4\left( \frac{p}{3} \right)^3}}{2} = -\frac{q}{2} \pm \sqrt{\left( \frac{q}{2} \right)^2 + \left( \frac{p}{3} \right)^3} \end{aligned}$$

In a similar manner, $$\begin{aligned} u = \frac{p}{3v} &\Longrightarrow q = v^3 - u^3 = v^3 - \left( \frac{p}{3v} \right)^3 \Longrightarrow v^3 - q - \left( \frac{p}{3v} \right)^3 = 0 \\\\ &\Longrightarrow v^6 - qv^3 - \left( \frac{p}{3} \right)^3 = 0 \\\\ &\Longrightarrow v^3 = \frac{q \pm \sqrt{q^2 + 4\left( \frac{p}{3} \right)^3}}{2} = \frac{q}{2} \pm \sqrt{\left( \frac{q}{2} \right)^2 + \left( \frac{p}{3} \right)^3} \end{aligned}$$

From now on, let $\Delta = \left( q/2 \right)^2 + \left( p/3 \right)^3$. $$\begin{aligned} u^3 = -\frac{q}{2} \pm \sqrt{\Delta}, \qquad v^3 = \frac{q}{2} \pm \sqrt{\Delta} \end{aligned}$$

Case 2. $\Delta = 0$ (All roots are real and two equal)

This case makes $u^3 = -q/2$ and $v^3 = q/2$. Recall that the solutions of $y^3 - 1 = 0$ are $y = 1, e^{2\pi i/3}, e^{4\pi i/3}$ according to the root of unity. Keeping in mind $p = 3uv$ is constant which means that $uv$ must be constant, it indicates that $u$ and $v$ are rotated in the opposite direction of each other in the complex plane. $$\begin{aligned} u &= \sqrt[3]{-\frac{q}{2}}, \quad \sqrt[3]{-\frac{q}{2}} e^{-2\pi i/3}, \quad \sqrt[3]{-\frac{q}{2}} e^{-4\pi i/3} \\\\ v &= \sqrt[3]{\frac{q}{2}}, \quad \sqrt[3]{\frac{q}{2}} e^{2\pi i/3}, \quad \sqrt[3]{\frac{q}{2}} e^{4\pi i/3} \end{aligned}$$

• If $u = \sqrt[3]{-q/2}$ and $v = \sqrt[3]{q/2}$, the solution of the reduced form is $t = u - v = 2\sqrt[3]{-q/2}$, which produces $x = 2\sqrt[3]{-q/2} - a/3$.

• If $u = \sqrt[3]{-q/2}e^{-2\pi i/3}$ and $v = \sqrt[3]{q/2}e^{2\pi i/3}$,

$$\begin{aligned} t &= u - v = \sqrt[3]{-\frac{q}{2}} e^{-2\pi i/3} - \sqrt[3]{\frac{q}{2}} e^{2\pi i/3} \\\\ &= -\sqrt[3]{\frac{q}{2}} \left( \cos \left( -\frac{2\pi}{3} \right) + i \sin \left( -\frac{2\pi}{3} \right) + \cos \left( \frac{2\pi}{3} \right) + i \sin \left( \frac{2\pi}{3} \right) \right) \\\\ &= -\sqrt[3]{\frac{q}{2}} \left( -1 \right) = \sqrt[3]{\frac{q}{2}} \end{aligned}$$

which produces $x = \sqrt[3]{q/2} - a/3$.

• If $u = \sqrt[3]{-q/2}e^{-4\pi i/3}$ and $v = \sqrt[3]{q/2}e^{4\pi i/3}$,

$$\begin{aligned} t &= u - v = \sqrt[3]{-\frac{q}{2}} e^{-4\pi i/3} - \sqrt[3]{\frac{q}{2}} e^{4\pi i/3} \\\\ &= -\sqrt[3]{\frac{q}{2}} \left( \cos \left( -\frac{4\pi}{3} \right) + i \sin \left( -\frac{4\pi}{3} \right) + \cos \left( \frac{4\pi}{3} \right) + i \sin \left( \frac{4\pi}{3} \right) \right) \\\\ &= -\sqrt[3]{\frac{q}{2}} \left( -1 \right) = \sqrt[3]{\frac{q}{2}} \end{aligned}$$

which also produces $x = \sqrt[3]{q/2} - a/3$.

It is noticeable that two equal real roots rather come out after considering all the imaginary parts.

Case 3. $\Delta > 0$ (One is a real root and two complex roots)

$$\begin{aligned} t &= u - v = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\Delta}} - \sqrt[3]{\frac{q}{2} \pm \sqrt{\Delta}} = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} \mp \sqrt{\Delta}} \\\\ &= \sqrt[3]{-\frac{q}{2} + \sqrt{\Delta}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\Delta}} \end{aligned}$$

Note that $\pm$ is eventually disappeared. So, $x = \sqrt[3]{-q/2 + \sqrt{\Delta}} + \sqrt[3]{-q/2 - \sqrt{\Delta}} - a/3$.

Case 4. $\Delta < 0$ (All roots are real and different)

In this case, $u$ and $v$ are complex numbers. $$\begin{aligned} u^3 = -\frac{q}{2} \pm i \sqrt{-\Delta}, \qquad v^3 = \frac{q}{2} \pm i \sqrt{-\Delta} \end{aligned}$$

Recall that a complex number $m + ni$ can be represented as $r (\cos \phi + i \sin \phi)$ where $r = \sqrt{m^2 + n^2}$, $\cos \phi = m/r$, and $\sin \phi = n/r$. With this, $u^3$ can be rewritten as $r (\cos \phi + i \sin \phi)$ where $$\begin{aligned} r = \sqrt{\left( \frac{q}{2} \right)^2 - \Delta}, \qquad \cos \phi = -\frac{q}{2r}, \qquad \sin \phi = \frac{\pm \sqrt{-\Delta}}{r} \end{aligned}$$

Then, $v^3$ can be also rewritten, noting that it has the same $r$ of $u$. $$\begin{aligned} r_v = \sqrt{\left( \frac{-q}{2} \right)^2 - \Delta} = r, \qquad \cos \phi_v = \frac{q}{2r} = - \cos \phi, \qquad \sin \phi_v = \frac{\pm \sqrt{-\Delta}}{r} = \sin \phi \end{aligned}$$

which means that $v^3 = r(-\cos \phi + i \sin \phi) = -r(\cos \phi - i \sin \phi) = -r (\cos (-\phi) + i \sin (-\phi))$. More simply, $u^3 = r e^{i \phi}$ and $v^3 = -r e^{-i \phi}$. Again, keeping in mind $uv$ is constant, rotate each $u$ and $v$ in the opposite direction in the complex plane. $$\begin{aligned} u &= \sqrt[3]{r} e^{\phi i/3}, \quad \sqrt[3]{r} e^{(\phi + 2 \pi)i/3}, \quad \sqrt[3]{r} e^{(\phi + 4 \pi)i/3} \\\\ v &= \sqrt[3]{-r} e^{-\phi i/3}, \quad \sqrt[3]{-r} e^{-(\phi + 2 \pi)i/3}, \quad \sqrt[3]{-r} e^{-(\phi + 4 \pi)i/3} \end{aligned}$$

• If $u = \sqrt[3]{r} e^{\phi i/3}$ and $v = \sqrt[3]{-r} e^{-\phi i/3}$,

$$\begin{aligned} t &= u - v = \sqrt[3]{r} e^{\phi i/3} - \sqrt[3]{-r} e^{-\phi i/3} \\\\ &= \sqrt[3]{r} \left( \cos \left( \frac{\phi}{3} \right) + i \sin \left( \frac{\phi}{3} \right) + \cos \left( -\frac{\phi}{3} \right) + i \sin \left( -\frac{\phi}{3} \right) \right) \\\\ &= 2\sqrt[3]{r} \cos \left( \frac{\phi}{3} \right) \end{aligned}$$

which produces $x = 2\sqrt[3]{r} \cos \left( \phi/3 \right) - a/3$.

• If $u = \sqrt[3]{r} e^{(\phi + 2 \pi)i/3}$ and $v = \sqrt[3]{-r} e^{-(\phi + 2 \pi)i/3}$,

$$\begin{aligned} t &= u - v = \sqrt[3]{r} e^{(\phi + 2 \pi)i/3} - \sqrt[3]{-r} e^{-(\phi + 2 \pi)i/3} \\\\ &= \sqrt[3]{r} \left( \cos \left( \frac{\phi + 2 \pi}{3} \right) + i \sin \left( \frac{\phi + 2 \pi}{3} \right) + \cos \left( -\frac{\phi + 2 \pi}{3} \right) + i \sin \left( -\frac{\phi + 2 \pi}{3} \right) \right) \\\\ &= 2\sqrt[3]{r} \cos \left( \frac{\phi + 2 \pi}{3} \right) \end{aligned}$$

which produces $x = 2\sqrt[3]{r} \cos \left( (\phi + 2 \pi)/3 \right) - a/3$.

• If $u = \sqrt[3]{r} e^{(\phi + 4 \pi)i/3}$ and $v = \sqrt[3]{-r} e^{-(\phi + 4 \pi)i/3}$,

$$\begin{aligned} t &= u - v = \sqrt[3]{r} e^{(\phi + 4 \pi)i/3} - \sqrt[3]{-r} e^{-(\phi + 4 \pi)i/3} \\\\ &= \sqrt[3]{r} \left( \cos \left( \frac{\phi + 4 \pi}{3} \right) + i \sin \left( \frac{\phi + 4 \pi}{3} \right) + \cos \left( -\frac{\phi + 4 \pi}{3} \right) + i \sin \left( -\frac{\phi + 4 \pi}{3} \right) \right) \\\\ &= 2\sqrt[3]{r} \cos \left( \frac{\phi + 4 \pi}{3} \right) \end{aligned}$$

which produces $x = 2\sqrt[3]{r} \cos \left( (\phi + 4 \pi)/3 \right) - a/3$.

Alternative Form

Besides all $\cos$ forms as above, the alternative form can be derived with $\sin$ and a quadratic equation. First, let the last root above be $\alpha$. From $\cos \left( \pi/2 + \theta \right) = -\sin \theta$ and $\sin \theta = \sin \left( \pi - \theta \right)$, $$\begin{aligned} \alpha &= 2\sqrt[3]{r} \cos \left( \frac{\phi + 4 \pi}{3} \right) = -2\sqrt[3]{r} \sin \left( \frac{\phi}{3} + \frac{5 \pi}{6} \right) = -2\sqrt[3]{r} \sin \left( \pi - \left( \frac{\phi}{3} + \frac{5 \pi}{6} \right) \right) \\\\ &= -2\sqrt[3]{r} \sin \left( \frac{\pi}{6} - \frac{\phi}{3} \right) = -2\sqrt[3]{r} \sin \left( \frac{1}{3} \left( \frac{\pi}{2} - \phi \right) \right) \end{aligned}$$

Since $$\begin{aligned} \cos \phi = -\frac{q}{2r} = \sin \left( \frac{\pi}{2} - \phi \right) \end{aligned}$$

$\alpha$ can be rewritten only using $\sin$ and $\sin^{-1}$ forms. $$\begin{aligned} \alpha &= -2\sqrt[3]{r} \sin \left( \frac{1}{3} \left( \frac{\pi}{2} - \phi \right) \right) = -2\sqrt[3]{r} \sin \left( \frac{1}{3} \sin^{-1} \left( -\frac{q}{2r} \right) \right) \end{aligned}$$

Once $\alpha$ is obtained, the reduced cubic equation can be factorized. $$\begin{aligned} t^3 + pt + q = (t - \alpha) (t^2 + \alpha t + \alpha^2 + p) = 0 \end{aligned}$$

This implies that the other roots come from the quadratic equation $t^2 + \alpha t + \alpha^2 + p$ which is relatively easy work. $$\begin{aligned} t = \frac{-\alpha \pm \sqrt{\alpha^2 - 4(\alpha^2 + p)}}{2} = \frac{-\alpha \pm \sqrt{-3\alpha^2 - 4p}}{2} \end{aligned}$$

To sum up, all solultions are $$\begin{aligned} x &= -2\sqrt[3]{r} \sin \left( \frac{1}{3} \sin^{-1} \left( -\frac{q}{2r} \right) \right) - \frac{a}{3} = \alpha - \frac{a}{3} \\\\ x &= \frac{-\alpha + \sqrt{-3\alpha^2 - 4p}}{2} - \frac{a}{3} \\\\ x &= \frac{-\alpha - \sqrt{-3\alpha^2 - 4p}}{2} - \frac{a}{3} \end{aligned}$$

References

[1] Solving the Cubic Equation (Algebra)

[2] Cubic Equation Solver II