Complex solutions to $ x^3 + 512 = 0 $ An algebra book has the exercise 
$$ x^3 + 512 = 0 $$
I can find the real solution easily enough with 
$$ x^3 = -512 $$
$$ \sqrt[3]{x^3} = \sqrt[3]{-512} $$
$$ x = -8 $$
The book also gives the complex solutions $$ 4 \pm 4\sqrt{3}i $$
But I don't understand how to find these answers. Having completed the chapter on complex numbers I can find square roots of negative numbers easily, but cube (or higher) roots are never explained.

Hint: your full solution is $x=-8z$ where $z^3=1$. If you write $z=re^{i\theta}$ ($r\in\mathbb{R})$, then $z^3=r^3e^{i3\theta}=r^3(\cos(3\theta)+i\sin(3\theta))$. What can you say about $r$ and $\theta$ given that $z^3=1$?
