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.

With the information given, you could actually simplify the powers of $ab$ in succession
$(ab)^2 = b^4a\cdot ab = b^5$
$(ab)^3 = ab\cdot b^5 = a$
$(ab)^4 = a\cdot ab = b$
$(ab)^5 = ab\cdot b = ab^2$
Also $(ab)^5 = b\cdot b^4a = b^5a$
And finally  $(ab)^6 = b^5a\cdot ab = b^6=e$ meaning that $(ab)^{-1} = (ab)^5$
[However, also $(ab)^3 = b^5\cdot b^4a = b^3a$ which gives us $b^3=e$, but using that might confuse whoever is marking the question]   
