Find all complex numbers satisfying z^6=1

Hi guys.

"Find all complex numbers satisfying $\displaystyle z^6=1$"

The hint is to factorise $\displaystyle z^6-1$ as a difference of squares.

$\displaystyle (z^3+1)(z^3-1)$

$\displaystyle z(z^2+\frac{1}{z}) \cdot z(z^2-\frac{1}{z})$

This seems incorrect.

The bracketed terms are not quadratics of standard form and two z terms are identical, leading only to five solutions, from what I can see.

I've seen a solution involving De Moivre's theorem, in which a full rotation is split into six pieces, taking the $\displaystyle cis\Theta$ where $\displaystyle \Theta=0,\frac{\pi}{3}, \frac{2\pi}{3}, \pi , ..., 2\pi$.

I'm very interested in understanding the solution with De Moivre's theorem. It was explained in a course I took last semester but I simply cannot remember a thing about it.

I've asked several people about this problem, but they can't seem to distinguish copying from understanding.

I would love to really understand the solution to this problem.

Can you help?