Prove that if $a^{2} + ab + b^{2} | ab(a+b)$ then $\left | a-b \right | > \sqrt[3]{ab}$ This is one of my university homework's problems this week: 
Suppose that a and b are two positive integer and $a^{2} + ab + b^{2} | ab(a+b)$ then prove that $\left | a-b \right | > \sqrt[3]{ab}$
I tried to prove that $a^{2} + ab + b^{2} | (a-b)^{3}$ so I can say $a^{2} + ab + b^{2} \leq (a-b)^{3} \Rightarrow ab < (a-b)^{3}$ , but I'm stuck.
sorry for my poor English.  

We need $a \neq b$.
WLOG assume $a>b$. Let $d=(a,b)$. Then there exists two coprime positive integers $x> y$ such that $a=dx,b=dy$.
$a^{2} + ab + b^{2} | ab(a+b)$ equivalent to $x^2+xy+y^2 \mid dxy(x+y)$
Since $(x,y)=1$, $x^2+xy+y^2$ and $xy(x+y)$ are coprime. So we must have $x^2+y^2+xy \mid d$. Then $d\geq x^2+y^2+xy>xy$.
$(a-b)^3 > ab \iff d(x-y)^3>xy$. Since $x > y$ then $x-y \geq 1$, combine with above inequality gives the result.
