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?

We could look at it like this:
$$8^{\frac13}=2.1^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi}{n}\right)$$
Now for different values of $k$, we have different answers: (here  $n$ is $3$)
$$k=1\implies 8^{\frac13}=2\cdot\text{CiS} \left(\frac{2\pi}{3}\right)$$
$$k=2\implies8^{\frac13}=2\cdot\text{CiS}\left(\frac{4\pi}{3}\right)$$
$$k=3\implies8^{\frac13}=2\cdot\text{CiS}(2\pi)=2$$
You could read up on $n^{\text{th}}$ roots of unity on Wikipedia to get a better picture
