Suppose $A=\langle a,b\mid ab^2a^{-1}b^{-3},ba^2b^{-1}a^{-3}\rangle$, where $\langle a_1,\ldots, a_n \mid R\rangle$ is the group generated by $a_1,\ldots, a_n$ with relations in $R$. Show that $A \cong \{1\}$.
After carrying out some algebras, I obtain $ab^2=b^3a$ and $ba^2=a^3b$. From here I don't know how to proceed.
Can anyone give some hints?
$ab^2a^{-1}=b^3 \Rightarrow ab^4a^{-1}=b^6 \Rightarrow a^2b^4a^{-2}=ab^6a^{-1} = b^9$.
Now $a^2 = b^{-1}a^3b$, and substituting for $a^2$ in the equation $a^2b^4a^{-2}=b^9$ gives $a^3b^4a^{-3}=b^9$.
But $a^3b^4a^{-3}=b^9 = a(a^2b^4a^{-2})a^{-1} = ab^9a^{-1}$, so $ab^9a^{-1}=b^9$.
But $a^{-1}b^9a=b^6$, so $b^9=b^6$ and $b^3=1$, from which it is easy to prove that $a=b=1$.