TITLE: Prove that $ab$ is perfect cube.
QUESTION [6 upvotes]: Let $a,b$ be positive integers, $b<a$, such that $a^3+b^3+ab$ is divisible by $ab(a-b)$. Prove that $ab$ is perfect cube.
My try:
Let $a=dm$ and $b=dn$ such that $gcd(m,n)=1$. now $ab|(a^3+b^3+ab)$ so $ab|a^3+b^3$.
now, $\frac{d(m^3+n^3)}{(mn)}$ is an integer so both $m$ and $n$ divide $d$ since $gcd(m,n)=1$. Let $d=mnk$. Hence $a=m^2nk$ and $b=mn^2k$ so $ab=m^3n^3k^2$. finally we need to prove that $k$ is a perfect cube. I am not getting this. Please help.

REPLY [6 votes]: You were so close . The problem is that you didn't used the whole condition :
$$ab(a-b) \mid a^3+b^3+ab$$
You only used the fact that $ab \mid a^3+b^3+ab$.
We know that $a=dm=km^2n$ and $b=kn^2m$ so the condition is now :
$$k^3m^4n^4(m-n) \mid k^3m^6n^3+k^3n^6m^3+k^2m^3n^3$$ and now simplify it :
$$kmn(m-n) \mid km^3+kn^3+1$$but this means that : $$k \mid km^3+kn^3+1$$ or 
$$k \mid 1$$
This proves that $k=1$ and then : $$ab=(mn)^3$$ is a perfect cube .