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.

Solutions of any equation of the form $x^3=a^3$ are $a$, $a\omega$, $a\omega^2$, where $\omega$ is a cube root of unity, and $\omega \neq 1$. Usually one picks $\omega=e^{2i\pi/3}$
