<s>[INST] How to solve $x^3=-1$? How to solve $x^3=-1$? I got following: 
$x^3=-1$ 
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
 [/INST] Set $\displaystyle x=re^{i \theta}$. So $\displaystyle r^3e^{i3\theta}= x^3= -1= e^{i \pi}$, hence $r^3=1$ and $3 \theta= \pi [2\pi]$. Finally, $r=1$ and $\displaystyle \theta = \frac{\pi}{3} \left[\frac{2\pi}{3} \right]$ so you get three solutions: $$x_1= e^{i \pi/3}= \frac{1}{2}+ i \frac{\sqrt{3}}{2}, \ x_2=e^{-i 2 \pi /3}= \frac{1}{2} - i \frac{\sqrt{3}}{2}, \ x_3= e^{i \pi}=-1.$$
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