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

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.

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.