I used the method shown in the link (the second answer) to solve $0 = x^3 - 2x - 4$:
Is there a systematic way of solving cubic equations?
I got that one solution is $x = (2 + \frac{10}{3\sqrt{3}})^{1/3} + \frac{2}{3} (2 + \frac{10}{3\sqrt{3}})^{-1/3}$.
Is there are a way of showing that the above expression simplifies to 2?
$\displaystyle 2\pm\frac{10}{3\sqrt{3}}=\frac{\pm10+6\sqrt{3}}{3\sqrt{3}}=\frac{(\pm1)^3+3(\pm1)^2(\sqrt{3})+3(\pm1)(\sqrt{3})^2+(\sqrt{3})^3}{(\sqrt{3})^3}=\left(\frac{\pm1+\sqrt{3}}{\sqrt{3}}\right)^3$
\begin{align*} \left(2\pm\frac{10}{3\sqrt{3}}\right)^{1/3}+\frac{2}{3}\left(2\pm\frac{10}{3\sqrt{3}}\right)^{-1/3}&=\frac{\pm1+\sqrt{3}}{\sqrt{3}}+\frac{2}{3}\left(\frac{\sqrt{3}}{\pm1+\sqrt{3}}\right)\\ &=\frac{\pm1+\sqrt{3}}{\sqrt{3}}+\frac{2}{3}\left(\frac{\sqrt{3}(\sqrt{3}\mp1)}{3-1}\right)\\ &=2 \end{align*}
Indeed, you can solve the equation by factorization.
$$x^3-2x-4=(x^3-8)-2(x-2)=(x-2)(x^2+2x+4)-2(x-2)=(x-2)(x^2+2x+2)$$
Let $$x=\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}+\frac{2}{3\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}}$$
$$x^3=\left(\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}+\frac{2}{3\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}}\right)^3$$
$$x^3=2\pm \frac{10}{3\sqrt{3}}+\frac{8}{27\left(2\pm \frac{10}{3\sqrt{3}}\right)}+3x\left(\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}\cdot \frac{2}{3\sqrt[3]{2\pm \frac{10}{3\sqrt{3}}}}\right)$$
$$x^3=4+2x\rightarrow (x-2)(x^2+2x+2)=0$$
It's not hard to see $x^2+2x+2>0\rightarrow x=2$
Q.E.D
