Find ${(ab)}^{-1}$ if $a^2=e, b^6=e, ab=b^4a$ where $a,b$ are elements of a group 
If $a,b$ are elements of a group and $a^2=e, b^6=e, ab=b^4a$, then find the order of $ab$ and express ${(ab)}^{-1}$ in terms of $a^mb^n$ and $b^ma^n$ 

I could find the order of $ab$ to be 6 but struggling to find ${(ab)}^{-1}$ in terms of $a^mb^n$ and $b^ma^n$.
Please help me move forward.

We have 
$$b = a b^4 a = (a b a)^4= (b^4)^4=b^{16}= b^6 \cdot b^6 \cdot b^4 = b^4$$
so $b^3=e$ and $ab= b a$. From here it is easy to see that the order of $ab$ is $\le 6$  (it is in fact the product of the orders of $a$, $b$, these being relatively prime). Moreover, $(ab)^{-1}=a^{-1}b^{-1}=a b^2= b^2 a$.
