Let $G$ be a group and $a,b\in G$. Suppose that $$a^{-1}b^2a=b^3$$ and $$b^{-1}a^2b=a^3$$ Show that $a=b=1$.
If we can show that $a=b$, then from the two relations clearly $a=b=1$.
I observe that each relation can be rewritten as $[a^{-1},b^2]=b$ and $[b^{-1},a^2]=a$.
Also they can be written as $(a^{-1}ba)^2=b^3$ and $(b^{-1}ab)^2=a^3$.
I need some guidance to solve this kind of problem. I did some guess but they did not lead to the result.
