How do I find $\sqrt[3]{-i}$? I'm asked to evaluate $$\sqrt[3]{-i}$$
I suppose $\sqrt[3]{-i}=(a+bi)$
$$\implies (a+bi)^3=-i$$
$$\implies \Im \left( (a+bi)^3 \right) =\Im \left( (-i) \right)$$
$$\implies 3a^2b-b^3=-1$$
Now how am I supposed to find $a$,$b$? 
Aren't there infinitely of them instead of just three?
 A: Let us start as you did:
$\sqrt [3] {-i}= a+bi$
$-i=(a+bi)^3$
$-i=a^3+3a^2bi-3ab^2-b^3i$  
Therefore we get $2$ equations:
$a^3-3ab^2=0   \ \ \ \ldots(1)$
$3a^2b-b^3=-1 \ldots (2)$  
Solving for $(1)$:
$a(a^2-3b^2)=0$
$\therefore a(a-b\sqrt3)(a+b\sqrt 3)=0$  
So:
$a=0$
$a=b\sqrt3$
$a=-b\sqrt 3$  
Solving for $(2)$:
$3a^2b-b^3=-1$
When $a=0$:
$-b^3=-1$
$b^3=1$
$b=1$  
When $a=\sqrt {3b^2}$:
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=\frac{\sqrt 3}2$
When $a=-\sqrt {3b^2}$
$9b^3-b^3=-1$
$b^3=-\frac 18$
$b=-\frac 12$
$\therefore a=-\frac{\sqrt 3}2$  

So your $3$ solutions are:
  $(a,b)=(0,1);(\frac{\sqrt 3}2,-\frac 12);(-\frac{\sqrt 3}2,-\frac 12)$  

I sincerely hope this helps you to understand why there are only $3$ solutions without mentioning the fundamental theorem of algebra. Good luck!  
A: You can solve this as 
\begin{equation}
 (-i)^{\frac{1}{3}} 
 =
 ( e^{i \frac{3\pi}{2} + i2k\pi})^{\frac{1}{3}} 
\end{equation}
which gives us three distinct roots $e^{i z_k}$, for $k = 0,1,2$, where
\begin{align}
 z_0 &=  \frac{1}{3}\frac{3\pi}{2} + \frac{2(0)\pi}{3} = \frac{\pi}{2} \\
 z_1 &=  \frac{1}{3}\frac{3\pi}{2} + \frac{2(1)\pi}{3} = \frac{7\pi}{6} \\
 z_2 &=  \frac{1}{3}\frac{3\pi}{2} + \frac{2(2)\pi}{3} = \frac{11\pi}{6} \\
\end{align}
If you insist on solving it your way, then 
\begin{equation}
 (a+bi)^3 = -i
\end{equation}
means 
\begin{equation}
 a^3 + 3a^2bi - 3ab^2 - b^3i = -i
\end{equation}
which means
\begin{align}
 a^3 - 3ab^2 &= 0 \\
 3a^2b - b^3 &= -1
\end{align}
which is
\begin{align}
 a^2 - 3b^2 &= 0 \\
 3a^2b - b^3 &= -1
\end{align}
or
\begin{align}
 (a - \sqrt{3} b)(a + \sqrt{3} b) &= 0 \\
 3a^2b - b^3 &= -1
\end{align}
The first equation suggests either
\begin{equation}
 a = \pm \sqrt{3} b
\end{equation}
Replacing this in the second equation gives
\begin{equation}
 3(\sqrt{3} b)^2b - b^3 = -1
\end{equation}
which is
\begin{equation}
 9b^3 - b^3 = -1
\end{equation}
i.e.
\begin{equation}
 b = -\frac{1}{2}
\end{equation}
This will give us 
\begin{equation}
 a = \pm \sqrt{3} (-\frac{1}{2})
\end{equation}
which means we get two solutions
\begin{align}
 (a_1,b_1) &= (-\sqrt{3},-\frac{1}{2}) \\
 (a_2,b_2) &= (\sqrt{3},-\frac{1}{2}) \\
\end{align}
which are actually what we found before, i.e. 
\begin{align}
 a_1 + ib_1 &= e^{i z_1} \\
 a_2 + ib_2 &= e^{i z_2} \\
\end{align}
Then you've got one more root you need to find (which is the obvious one), where $a_0 = 0$ and $b_0 = 1$, i.e. $i^3 = -i$. Also, you can verify that
\begin{align}
 a_0 + ib_0 &= e^{i z_0} 
\end{align}
A: When you take the real part in each side, you get another equation. I don't think that is the best way to tackle the problem though. I think it is better:
$$z^3=-i$$
Then we know the magnitude of $z$ is $1$, and the angle is such that when multiplied by 3 it lies at $\frac{3\pi}{2}+2k\pi, k\in \text{Z}$ (this is pointing downwards). The possible angles are:
$$\frac{\pi}{2}+\frac{2k\pi}{3}$$
So take the cases $k=0, k=1 $ and $k=2$ and you are done, because then the angles start to loop (just adding $2\pi$). The solutions are:
$$z=\cos{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)} + i\sin{\left(\frac{\pi}{2}+\frac{2k\pi}{3}\right)}, k\in\text{{0,1,2}}$$
A: Alt. hint:   the cube roots of $\,-i\,$ are the solutions to $\,z^3=-i \iff z^3+i=0\,$. Using that $\,i = -i^3\,$ and the identity $\,a^3-b^3=(a-b)(a^2+ab+b^2)\,$, the latter equation can be written as:
$$
0 = z^3+i = z^3-i^3=(z-i)(z^2+iz + i^2)=(z-i)\big(-(iz)^2+iz-1\big)
$$
The first factor gives the (obvious) solution $\,z=i\,$, and the second factor is a real quadratic in $\,iz\,$ with roots $\,iz = (1 \pm i \sqrt{3})/2 \iff z = \ldots$
A: It is easier to find the cube roots of -i using the polar form.
Note that $-i = e^{3i\pi /2}$ thus  the cube roots are $e^{i\pi/2},e^{i\pi/2+2i\pi/3},  e^{i\pi/2+4i\pi/3}$ 
You can easily find these roots in $a+bi $ form if you wish to do so.
