I would like to solve the equation $$x^3=(1256)(37)$$ in $S_7$.
I went like this.
Let's consider $x$'s decomposition into disjoint cycles. $x$ cannot have a cycle of length $5$ or of length $7$ in its decomposition, because these cubed result in a cycle of length $5$ or of length $7$.
Also, a cycle of length $6$ cubed gives us a product of $3$ disjoint cycles of length $2$ and this doesn't work either. So, $x$ can only be the product of disjoint cycles of length $2$, $3$ and $4$.
Now, a cycle of length $2$ cubed equals itself, a cycle of length $3$ cubed is the identity and a cycle of length $4$ cubed is a cycle of length $4$. Since $x^3$ must have a cycle of length $4$ and one of length $2$ in its decomposition, we see that it is necessary that $x$ be the product of a cycle of length $2$ and a cycle of length $4$ (there is no room for some other cycle of length $3$).
So, after some computations, we get that the only solution is $x=(37)(1652)$. Is this solution correct?