Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ The question is:
Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. 2
b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)
c. 2, 2 cis ($\pi$/3)
d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)
e. None of these

I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$
And, $r = 8$
And, $8\cos \theta = 8$ and $\theta = 0$.
So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$
I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?

$8^{\frac{1}{3}}$=$2(1)^{\frac{1}{3}}=2,2\omega,2{\omega}^2$
here $\omega$ is cube root of unity
