Solution:
\( \displaylines{z=8e^{\frac{i3\pi}{2}},\text{ taking the cubed root of this gives us a principal value }z^{\frac13}=2e^{\frac{i\pi}{2}}=2i\\ \text{Multiplying by }w=e^{i\frac{2\pi}{3}}\text{ or }w^2=e^{i\frac{4\pi}{3}}\text{ will yield the other two cube roots, which work out to be}\sqrt3\pm i} \)