Solve $x^2=(12)(34)$ and $x^3=(12)(34)$ in $S_4$

Question

Solve the following equations in $S_4$:
a)$x^2=(12)(34)$
b)$x^3=(12)(34)$
I know that a) appears here Find $\alpha \in S_4$ such that $\alpha^2=(12)(34)$, but I want to present my reasoning to see if I know how to tackle such problems correctly.
a) Let $x=\pi_1...\pi_k$ be $x$'s decomposition into disjoint cycles.
Since $\operatorname{ord}(x)=4$, these cycles must be of length $2$ and $4$. When we raise $x$ to the second power, the transpositions equal $e$ and we are left only with the squares of the cycles of length $4$. But $(a b c d)^2=(a c)(b d)$. It now follows that $x$ must be a cycle of order $4$, since the decomposition into disjoint cycles is unique. Now it is easy to find that we have $x\in \{(1324),(2314)\}$.
b) Again, let $x=\pi_1...\pi_k$ be $x$'s decomposition into disjoint cycles.
Since $\operatorname{ord}(x)=6$, these cycles must be of length $2$, $3$ or $6$. The transpositions to the third power are equal to themselves, the cycles of length $3$ are equal to $e$ and the cycles of length $6$ turn into $3$ disjoint tranpositions. Now, since $x^3$ decomposes into two disjoint transpositions, it follows that $x$ can't have any cycles of length $6$ in its decomposition and it also can't have any cycles of length $3$ either (if it had, $(12)(34)$ also would). Hence, $x=(12)(34)$ is the only solution.