For give permutation $\sigma\in S_{13}$ solve equation $x^3=\sigma$ We have this permutation 
$$\sigma =\left({\begin{array}{*{20}c}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\ 3 & 2 & 1 & 11 & 5 & 8 & 13 & 10 & 9 & 12 & 4 & 6 & 7\end{array}}\right)\ $$
Find $x^3=\sigma\in S_{13}$
The cycles are (1 3)(2)(4 11)(5)(6 8 10 12)(9)(13 7)

The permutation with the cycles $(1\ 3)(2)(4\ 11)(5)(6\ 12\ 10\ 8)(9)(13\ 7)$ does the job.
You only have to invert the $4$-cycle. The other cycles remain. You can also take the cycle $(2\ 5\ 9)$ to get another solution.
