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.
 A: As indicated in hints beneath the OP,
$$(ab)^{-1}=b^{-1}a^{-1}=b^5a$$
which gives the inverse in the form $b^ma^n$.  To get it in the form $a^mb^n$, use the equation $ab=b^4a$ to obtain
$$(ab)^{-1}=(b^4a)^{-1}=a^{-1}b^{-4}=ab^2$$
Added later: It's worth noting that $(ab)^{-1}=ab^2=b^5a$ implies $(ab)^{-2}=(ab^2)(b^5a)=aba$ and hence (using what the OP found, namely $(ab)^6=e$)
$$e=(ab)^{-6}=(aba)(aba)(aba)=ab^3a$$
from which it follows that $b^3=e$, so the order of $b$ is not $6$. (This was found by orangeskid in more direct fashion.)
A: 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, $(ab)^{-1}=a^{-1}b^{-1}=a b^2= b^2 a$.
A: 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]   
