How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .

By the fundamental theorem of algebra, we expect there to be 3 solutions.
If $z$ is a solution to $z^3 = -8$, then $|z^3| = |-8| = 8$, but since $|ab| = |a| * |b|$ and $|z|$ is a real number $\geq 0$, this implies $|z|^3 =8$ so $|z| = 2$.
Therefore we have the magnitude of all solutions to $z^3 = -8$. Now, we just need to match the angle. We have $(r e^{i \theta})^3 = r^3 e^{3 i \theta}$, and we already know that $r = 2$, so we just need $3 \theta = \pi (\mod 2 \pi)$.
$\theta = \pi/3$ is obviously a solution, for the other two solutions, add $2 \pi/3$ and $4 \pi/3$.
Therefore, the solutions are $2 e^{i \pi/3}, 2 e^{3 \pi i / 3} = -2$ and  $2 e^{5 \pi i/3}$.
