Solve cubic equation $x^3 - 15x - 4 = 0$ with Cardano's method

Question

Suppose that we wanted to solve $$x^3 - 15x - 4 = 0$$ by using Cardano's method. In doing so, I end up with $$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$ However, by checking my answer, it seems there are two other answers I haven't accounted for: $\sqrt{3} - 2$ and $-2 - \sqrt{3}$. Is there a way to generate these solutions from this method, or would I need to find them in other ways, e.g., by factoring?

Answer 1

Yes, there is. First of all, note that Cardano's formula doesn't say that the roots are the sum of any cubic root of $2+11i$ with any cubic root of $2-11i$. It says that they are the sum of a cubic root $u$ of $2+11i$ with a cubic root $v$ of $2-11i$ such that $uv=-\frac{-15}3=5$.

Now, take $u=2+i$ and $v=2-i$. Then indeed $u^3=2+11i$, $v^3=2-11i$, and $uv=5$. Therefore, $4(=u+v)$ is a root.

But now take $\omega=-\frac12+\frac{\sqrt3}2i$. Then $\omega^2=-\frac12-\frac{\sqrt3}2i$, $\omega^3=1$ and if you take $u'=\omega u$, $v'=\omega^2v$, $u''=\omega^2u$, and $v''=\omega v$, then $u'+v'$ and $u''+v''$ will be roots too. And that's how you get $-2\pm\sqrt3$: $u'+v'=-2-\sqrt3$ and $u''+v''=-2+\sqrt3$.

Answer 2

The Cardano’s formula is normally used to calculate the real root of the a cubic equation with one real root and a pair of complex roots. In the case of three real roots, e.g. $x^3 - 15x - 4 = 0$, the formula results in the expression

$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$

which is of multiple complex values. However, it can still be used to obtain the three real roots if their real parts are taken, i.e.

$$x=Re(\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i})= 2Re (\sqrt[3]{2 + 11i})\tag1 $$ Write the cubic radicant as $$2+11i=\sqrt{125}e^{i \cos^{-1}\frac2{5\sqrt5}} = \sqrt{125}e^{i \theta}$$ where $$\cos\theta = \frac2{5\sqrt5}=4\cos^3\frac{\theta}3-3\cos\frac{\theta}3 $$ which factorizes as $$\left(\cos\frac{\theta}3- \frac2{\sqrt5}\right)\left(\cos\frac{\theta}3+\frac{\sqrt3+2}{\sqrt5}\right)\left(\cos\frac{\theta}3- \frac{\sqrt3-2}{\sqrt5}\right)=0 $$ Then $$\cos\frac{\theta}3=\frac2{\sqrt5},\>\frac{\sqrt3-2}{2\sqrt5},\> \frac{-\sqrt3-2}{2\sqrt5} $$ and the corresponding roots (1) are \begin{align} x& =2Re \left( \sqrt{125}e^{i \theta}\right)^{\frac13} = 2\sqrt5 \cos\frac{\theta}3=4,\>\sqrt3-2,\>-\sqrt3-2 \end{align}