If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$. Suppose $a$ and $b$ are elements of the group $G$. If $a^2=e$ and $a^{-1}b^2a=b^3$, prove $b^5=e$.
I'm trying to prove it as follow.
Here's my solution:
If $a^2=e$ so $a=a^{-1}$. So we have $b^3=ab^2a$.
$$b^2=ab^3a$$
So I thought if I can show that $b^3b^2=b^2 b^3=e$ I can prove it.
$$ab^2aab^3a=ab^5a$$ so I'm thinking something here. we get $b^5=ab^5a$.

We are given that $b^3 = ab^2a$. If we cube this equation, we obtain that $b^9 = ab^6a = ab^3 \cdot b^3a$. Since from the given equation we deduce that $ab^3 = b^2a$ and $b^3a = ab^2$, we can make these substitutions to obtain $b^9 = b^2a \cdot ab^2 = b^4$. Hence, $b^5 = 1$, as desired.
Edit: The idea above shows that if $b^n$ and $b^m$ are conjugated in a group by an element of order $2$, then $b^{n^2-m^2} = 1$, which is a pretty neat and fairly general result.
