Question

Find the cube roots of 8(cos 216° + i sin 216°).

Answer 1

$z^3=8(\cos216^\circ+i\sin216^\circ)$
$z^3=2^3(\cos(6^3)^\circ+i\sin(6^3)^\circ)$
$\implies z=8^{1/3}\left(\cos\left(\dfrac{216+360k}3\right)^\circ+i\sin\left(\dfrac{216+360k}3\right)^\circ\right)$

where $k=0,1,2$. So the third roots are

$z=\begin{cases}2(\cos72^\circ+i\sin72^\circ)\\2(\cos192^\circ+i\sin192^\circ)\\2(\cos312^\circ+i\sin312^\circ)\end{cases}$