If $G$ is a group, show that $x^2ax=a^{-1}$ has a solution if and only if $a$ is a cube in $G$

Question

I was checking my old set of homework problems that I found this one:

If $G$ is a group, show that $x^2ax=a^{-1}$ has a solution if and only if $a$ is a cube in $G$.

One direction is easy. If $a$ is a cube in $G$ then there exists $y \in G$ such that $a=y^3$. Now with direct manipulations it is shown that $x=y^{-2}$ is a solution to $x^2ax=a^{-1}$.

The other direction is harder though. I first attempted to isolate $x$ and then somehow show that it's necessary for $a$ to be a cube. But since I had no idea how to isolate $x$ I gave up. Then I decided to prove the existence of $y \in G: a=y^3$ by showing that a particular map is bijective so I can come up with a nice variable change to conclude that $a$ must be a cube, but this attempt failed as well. I'm tired now, so I think I'm giving up. I'm looking for a hint or a complete solution if a hint is not possible at this stage.

Answer 1

Starting with $x^{2}ax=a^{-1}$, try to manipulate the left-hand side to $xaxaxa= (xa)^3$ by judicious multiplications of the equation by $x, x^{-1}, a$, and $a^{-1}$. You will be pleasantly surprised by what ends up on the right.

Answer 2

Take your solution to the first direction as a clue: if $a = y^3$ and $x = y^{-2}$, then $y = ax = xa$. So it might be natural to guess that $a$ will be equal to $y^3$ for this choice of $y$, and indeed, it is not difficult to show that this is the case.