I want to show $$\langle a,b\mid ab^2＝b^3a, ba^2＝a^3\rangle$$ is isomorphic to a trivial group.
My text book proved this like following.
$bab^2a＝a^4b^2a\to ba＝a^4b^2$
$a^4b^2a＝a^3b$ (omitted some process), so $ab^2a＝b$
$b^2a^2＝1, a＝a^4b^3a b^3a^2＝1, b^3＝b^2\to b＝1, a＝1$
This proof is too difficult for me to imitate because there is no oriented strategy of each process. Firstly, I cannot understand why we should start from $bab^2a$.
Other way to prove is also appreciated, I would be happy if you could give me an strategy of deformation.
One way to see that $\langle a,b\mid ab^2＝b^3a, ba^2＝a^3\rangle$ defines the trivial group is to use the second relator as follows: $$\begin{align*} ba^2&=a^3\\ \Rightarrow ba^2\cdot a^{-2}&=a^3\cdot a^{-2}\\ \Rightarrow b&=a \end{align*}$$ Then rewrite $b\rightarrow a$ in the first relator: $$\begin{align*} ab^2&=b^3a\\ \Rightarrow a^3&=a^4\\ \Rightarrow a&=1, b=1\end{align*}$$ Hence, the group is trivial.
Note that there is no general algorithm to decide whether or not a group is trivial, but if a group is trivial then there is a procedure which will prove this. See here for more details.