Finding the 'cube roots' of a permutation

Question

How do I find three elements $\sigma \in S_9$ such that $\sigma^3=(157)(283)(469)$? Since the three $3$-cycles in $\sigma^3$ are disjoint, $|\sigma^3|=\operatorname{lcm}(3,3,3)=3$. Then since $(\sigma^3)^3=\sigma^9=e$ we have that $|\sigma|=9$, so $\sigma$ is a $9$-cycle. That is all I'm really able to say.

How do we intelligently/methodically find the desired elements? A generalized approach is welcome as well.

Thanks.

Answer 1

Hint: One is $\sigma=(124\,586\,739)$.

Answer 2

We may assume $\sigma=(1ab5cd7ef)$. Now $2$ can be any of $a,b,c,d,e,f$. Let's say $a=2$. Then we have $\sigma=(12b58d73f)$. Now $4$ can be any of $b,d,f$.