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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

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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 $\sqrt[3]{2}$.

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Hint: $x^3-2=(x-2^{\frac{1}{3}})(x^2+2^{\frac{1}{3}}x+2^{\frac{2}{3}})$.

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Let $x=z\sqrt[3]2$. Then $0=x^3-2=2z^3-2$ and so $z^3=1$. Write $z=e^{i\theta}$ and solve for $\theta$.

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