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?

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!  
