How to solve $x^3=-1$? How to solve $x^3=-1$? I got following: 
$x^3=-1$ 
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...

Let $x=a+bi$, where $a$ and $b$ are real.  Then $(a+bi)^3 = -1$.  Expanding the left-hand side gives $$a^3 +3a^2bi -3ab^2 -b^3i = -1.$$
We can separate the real and imaginary parts of this equation:
$$\begin{align}
a^3 -3ab^2 & = -1 \\
3a^2b - b^3 & = 0
\end{align}$$
Taking the obvious $b=0$ solution gives us $a=-1$ and thus the real solution

$$x =-1.$$

So suppose $b\ne 0$. Then the second equation has $3a^2 = b^2$, so $b = \pm a\sqrt3$. Putting $3a^2$ for $b^2$ in the first equation gives $$ a^3 - 9a^3 = -1$$ so $a = \frac12.$ Since $b = \pm a\sqrt3$, we have $b=\pm\frac{\sqrt3}2$. This gives us the other two solutions, namely 

$$x=\frac12 \pm\!\frac{\sqrt3}2i.$$

