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?

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}
