<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] $x^3+1=0\implies (x+1)(x^2-x+1)=0$
If $x+1=0,x=-1$
else $x^2-x+1=0$ then  $x=\frac{1±\sqrt{1-4}}{2\cdot1}=\frac{1±\sqrt3i}{2}$ using the well known Quadratic Formula. 
Alternatively, using Euler's identity  and Euler's formula like other two solutions,
  for $x^n=-1=e^{i(2m+1)\pi}$ as $e^{i\pi}=-1$ where $m$ any any integer, $n$ is a natural number .
We know $n$-degree equation has exactly $n$ roots, so the roots of $x^n+1=0$
are $e^{\frac{i(2m+1)\pi}{n}}=\cos\frac{(2m+1)\pi}{n}+i\sin \frac{(2m+1)\pi}{n}$ where  $m=0,1,2,...n-1$. It's just customary, not mandatory that we have defined $m$  to assume this range of values, in fact it can assume any $n$ in-congruent values (for example, consecutive values like $r,r+1,...,r+n-1,$  where r is any integer), the reason is explained below.
(1)Using Repeated Root Theorem, let $f(x)=x^n+R$ where $n>1$ and $R ≠ 0\implies x≠0$
So, $\frac{df}{dx}=nx^{n-1}, \frac{df}{dx}=0$ does not have any non-zero root.
Clearly, $f(x)=x^n+R$ can not have any repeated root unless $R=0$
(2)Let $e^{\frac{i(2a+1)\pi}{n}}=e^{\frac{i(2b+1)\pi}{n}}$
$\implies e^{\frac{2i(a-b)\pi}{n}}=1=e^{2k\pi i}$ where $k$ is any integer.
$\implies a-b=kn\implies a≡b\pmod n$, so any $n$ in-congruent values of $m$ will give us essentially the same set of n distinct solutions.
Here $n=3$ so, let's take $m=0,1,2$.
$m=0\implies  \cos\frac{\pi}{3}+i\sin \frac{\pi}{3}=\frac{1+i\sqrt3}{2} $
$m=1\implies  \cos \pi+i\sin\pi=-1 $
$m=2\implies  \cos\frac{5\pi}{3}+i\sin \frac{5\pi}{3}=\frac{1-i\sqrt3}{2} $
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