Q.Find the cube roots of $8$.

The three cube roots of $8$ are $2,\ -1+i\sqrt3,\ -1-i\sqrt3$.

Write $8 = 8(\cos0+i\sin0)$. The cube roots are

$8^{1/3}\!\left(\cos\tfrac{2k\pi}{3}+i\sin\tfrac{2k\pi}{3}\right),\ k=0,1,2$, and $8^{1/3}=2$.

$k=0:\ 2(\cos0+i\sin0)=2.$

$k=1:\ 2(\cos120^\circ+i\sin120^\circ)=2\!\left(-\tfrac12+i\tfrac{\sqrt3}{2}\right)=-1+i\sqrt3.$ …