Answer
$2\sqrt[3] 2(cos40^\circ+i\ sin40^\circ)$
$2\sqrt[3] \\ 2(cos160^\circ+i\ sin160^\circ)$
$2\sqrt[3] \\ 2(cos280^\circ+i\ sin280^\circ)$
See graph.
Work Step by Step
1. Convert the complex number: $z=-8+8\sqrt 3i=16(-\frac{1}{2}+\frac{\sqrt 3}{2}i) =16(cos120^\circ+i\ sin120^\circ)$
2. We have: $z^{1/3}=2\sqrt[3] 2(cos(\frac{360k+120}{3})^\circ+i\ sin(\frac{360k+120}{3})^\circ)$
3. $k=0, z_0=2\sqrt[3] 2(cos40^\circ+i\ sin40^\circ)$
4. $k=1, z_1=2\sqrt[3] 2(cos160^\circ+i\ sin160^\circ)$
5. $k=2, z_2=2\sqrt[3] 2(cos280^\circ+i\ sin280^\circ)$
6. See graph.
