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?
 A: 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
A: Let $z^3=8$.
Thus, $$(z-2)(z^2+2z+4)=0,$$ which gives
$$\{2,-1+\sqrt3i,-1-\sqrt3i\}$$ or
$$\left\{2(\cos0+i\sin0),2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right), 2\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\right\}$$
A: Here,
$$\begin{align*}
8^{1/3} &= (|8|e^{2\pi kj})^\frac{1}{3}, k = 0,1,2\\
               &= |8|^\frac{1}{3} e^{\frac{2}{3}\pi kj},  k = 0,1,2\\
&= 2 e^{\frac{2}{3}\pi kj},  k = 0,1,2\\
\end{align*}$$
so,for $k=1$,$k=2$ we get $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$
Or take:
$$8^{1/3}=x$$
Then we get,
$$(x-2)(x^2+2x+4)=0$$
Then we get our desired roots.
A: $8^{\frac{1}{3}}$=$2(1)^{\frac{1}{3}}=2,2\omega,2{\omega}^2$
here $\omega$ is cube root of unity
