<s>[INST] 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?
 [/INST] 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.
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