polynomials - Find all roots of $x^{6} + 1$

I'm studying for my linear algebra exam and I came across this exercise that I can't solve.

Find all roots of polynomial $x^{6} + 1$. Hint: use De Moivre's formula.

I guessed that two roots are $i$ and $-i$, since:

$i^{6} = (i^{2})^{3} = (-1)^{3} = -1$

therefore, $i$ is root and his complex conjugate $-i$ has to be root too. However that was just guessing. I have no idea how can I use De Moivre's formula here.

Can you help me solve this?

Answer

Hint: if $x^6=-1$, then $|x|^6=1$ and you can write $x=\cos\theta + i\sin\theta$.

details:

Then the equation is, thanks to De Moivre theorem and
$\cos^2 + \sin^2 =1$, equivalent to
$$\cos 6\theta =-1\\ 6\theta = \pi\mod 2\pi\\ \theta\in \frac \pi 6+\left\{0, \frac\pi 3, \frac{2\pi}3,\pi,\frac{4\pi} 3, \frac{5\pi}3 \right\}.$$