<s>[INST] Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find  $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these

The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
 [/INST] Since $-8=8e^{i({\pi+2n\pi})}$ we have $(-8)^{\frac{1}{3}}=2e^{i\frac{\pi+2n\pi}{3}}$ for $n\in\mathbb Z$ (and you want the cases when $n=0,1$ and $2$).
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