Prove that $x$ has order $5$. let $ x \in G$ such that $(a^{-1})*(x^2)*(a) = x^3$ for some self inverse $a.$ Prove that $x$ has order $5.$
I don't know how to start this proof. Seems really difficult. 

$a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9$ but $x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9$ now using $a^2=a^{-2}=e$ we have $x^4=x^9$ so $x^5=e$ and because $5$ is prime $x=e$ or $x$ has order $5$.
