I am generalizing my question here about square roots in groups.
If I have a group $G$ with $|G|= 0 \mod 3$, then the map $f(x) = x^3$ is not one-to-one, and hence a cube root doesn't always exist.
If $|G| = 2 \mod 3$, then I can find the cube root of $g \in G$ by computing $g^{1/3} = g^{(|G|+1)/3}$.
The last case to consider is if $|G| = 1 \mod 3$. The previous technique will not work, but we can find for $g \in G$ an element $g^{2/3}$ such that $(g^{2/3})^3=g^2$ by choosing $g^{2/3} = g^{(|G|+2)/3}$.
If we also have $|G| = 1 \mod 2$, then by the previous question we can take a square root and hence get a cube root for each element. But what if $|G|=0 \mod 2$? Can we still always get a cube root?
Sure we can. If $|G|=1\mod 3$, then $g^{(2|G|+1)/3}$ is a cube root of $g$ for every $g\in G$.
This generalizes to $n$th roots. If $|G|=k\mod n$ and $k$ is relatively prime to $n$, then let $r$ be $k^{-1}\mod n$, so that $rk=1\mod n$. Then $g^{(-rk+1)/n}$ is an $n$th root of $g$ for all $g\in G$.