Let $(G, *)$ be a group and $a, b\in G$. Suppose that $ab^3a^{-1}=b^2$ and $b^{-1}a^2b=a^3$. Show that $a=b=e$.

Question

I was reading about group theory. There was a question given in the book :Let $(G, *)$ be a group and $a, b\in G$. Suppose that $ab^3a^{-1}=b^2$ and $b^{-1}a^2b=a^3$. Show that $a=b=e$.

What I can make out is only $ab^3=b^2a$ and $a^2b=ba^3$. Also may be, we can prove by induction $ab^{3n}a^{-1}=b^{2n}$ and $b^{-1}a^{2n}b=a^{3n}$...but what after that? I am not quite getting it...