Write z = -8 in polar form:
[tex]z = -8 = 8e^{i\pi}[/tex]
Then the cube roots of z are
[tex]z^{1/3} = 8^{1/3} e^{i\left(\frac{\pi+2n\pi}3\right)[/tex]
where n ∈ {0, 1, 2}, or
[tex]z^{1/3} \in \left\{8^{1/3} e^{i\pi/3}, 8^{1/3} e^{i\pi}, 8^{1/3} e^{i\,5\pi/3}\right\} \\\\ z^{1/3} \in \left\{2 \left(\dfrac12 + i\dfrac{\sqrt3}2\right), -2, 2 \left(\dfrac12-i\dfrac{\sqrt3}2\right)\right\} \\\\ \boxed{z^{1/3} \in \left\{1+i\sqrt3, -2, 1-i\sqrt3\right\}}[/tex]
