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

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}
