How can you find the cubed roots of $i$?

Using Euler's formula, which states $$ e^{i \theta} = \cos \theta + i \sin \theta $$ we will see that $$ i = 0 + i \cdot 1 = \cos \left( \frac{\pi}{2} + 2n \pi \right) + i \sin \left( \frac{\pi}{2} + 2n \pi \right) = e^{i \left(\frac{ \pi}{2} + 2n \pi \right)} $$ for all integers $n$. Thus, if $z^3 = i$, then $$z = \exp\left[ i \left(\frac{\pi}{6}+\frac{2n\pi}{3}\right)\right]$$ for all integers $n$.

I believe your "polynomial" approach would also have worked, if this is what you meant :

[In this, we are supposing that we knew nothing of the "Euler Identity", DeMoivre's Theorem, or roots of unity, all of which provide quite efficient devices]

If we (probably safely) assume that the solution(s) are complex numbers, and call $ \ z \ = \ a + bi \ , $ with $ \ a \ $ and $ \ b \ $ real, we can write the equation as

$$ (a + bi)^3 \ = \ a^3 \ + \ 3a^2 b \cdot i \ + \ 3a b^2 \cdot i^2 \ + \ b^3 i^3 \ = \ (a^3 \ - \ 3ab^2) \ + \ (3a^2b \ - \ b^3) \cdot i \ \ = \ \ i \ , $$

by applying the binomial theorem and "powers of $ \ i \ $ ". Since the right-hand side of the equation is a pure-imaginary number, this requires that

$$ a^3 \ - \ 3ab^2 \ = \ a \ ( a^2 \ - \ 3b^2 ) \ = \ 0 \ \ \text{and} \ \ 3a^2b \ - \ b^3 \ = \ b \ (3a^2 \ - \ b^2) \ = \ 1 \ \ . $$

The first equation presents us with two cases:

I -- $ \ a \ = \ 0 \ $ :

$$ a \ = \ 0 \ \ \Rightarrow \ \ b \ ( \ 0 \ - \ b^2 ) \ = \ -b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ -1 \ \ \Rightarrow \ \ z \ = \ 0 - i \ \ ; $$

II -- $ \ a^2 \ - \ 3b^2 \ = \ 0 $ :

$$ a^2 \ = \ 3b^2 \ \ \Rightarrow \ \ b \ ( \ 3 \cdot [3b^2] \ - \ b^2 \ ) \ = \ 8b^3 \ = \ 1 \ \ \Rightarrow \ \ b \ = \ \frac{1}{2} $$

$$ \Rightarrow \ \ a^2 \ = \ 3 \ \left( \frac{1}{2} \right)^2 \ = \ \frac{3}{4} \ \ \Rightarrow \ \ a \ = \ \pm \frac{\sqrt{3}}{2} \ \ \Rightarrow \ \ z \ = \ \frac{\sqrt{3}}{2} + \frac{1}{2}i \ , \ -\frac{\sqrt{3}}{2} + \frac{1}{2}i \ \ . $$

We have found three complex-number solutions to the equation. As Dan says, (one form of) the Fundamental Theorem of Algebra states that this third-degree polynomial with complex coefficients has, in all, three roots (counting multiplicities, which are each 1 here).

We probably wouldn't want to use this method for degrees higher than this, as the algebra would become more difficult to resolve. The techniques described by the other posters are far more generally used.

The answer of @Petaro is best, because it suggests how to deal with such questions generally, but here’s another approach to the specific question of what the cube roots of $i$ are.

You know that $(-i)^3=i$, and maybe you know that $\omega=(-1+i\sqrt3)/2$ is a cube root of $1$. So the cube roots of $i$ are the numbers $-i\omega^n$, $n=0,1,2$.