abstract algebra - How to find the roots of $x³-2$?

I'm trying to find the roots of $x^3 -2$, I know that one of the roots are $\sqrt[3] 2$ and $\sqrt[3] {2}e^{\frac{2\pi}{3}i}$ but I don't why.
The first one is easy to find, but the another two roots?

I need help

Thank you

Answer

If $\omega^3 = 1$ and $x^3 = 2$ then $(\omega x)^3 = \omega^3 x^3 = 2$.

Possible values of $\omega$ are $e^{\frac{1}{3}2 i \pi}$, $e^{\frac{2}{3}2 i \pi}$ and $e^{\frac{3}{3}2 i \pi}$. This is because $1 = e^{2 i \pi} = (e^{\frac{1}{k} 2 i \pi})^k$.

So the solutions of $x^3 - 2 = 0$ are $e^{\frac{1}{3}2 i \pi} \sqrt[3]{2}$, $e^{\frac{2}{3}2 i \pi} \sqrt[3]{2}$ and $\sqrt[3]{2}$.