# 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

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May have a look into: math.stackexchange.com/questions/192742/how-to-solve-x3-1 –  lab bhattacharjee Oct 25 '12 at 12:55
See here. –  Mhenni Benghorbal Oct 25 '12 at 12:58
There is a formula to solve cubic equations, but it is too complex to be true! It has an interesting history twist too. If you are interested, take a look at: en.wikipedia.org/wiki/Cubic_function –  Emmad Kareem Oct 25 '12 at 13:01

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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–  MvG Oct 25 '12 at 13:10

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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