Strange inequality in an NT problem If $a\neq b$ are positive integers and $a^2+ab+b^2 | ab(a+b)$ then show that $|a-b|>\sqrt[3]{ab}$.
WLOG $a>b$. $a^2+ab+b^2 | ab(a+b)-a(a^2+ab+b^2)=-a^3$, so we have $a^2+ab+b^2$ divides $a^3$ as well as $b^3$. From here I tried lots of things, but I don't think they are worth of mentioning. If you have any idea ,do not hesitate to suggest.

Notice that $$a^2+ab+b^2\mid ab(a+b)\iff a^2+ab+b^2\mid a^3\iff a^2+ab+b^2\mid b^3$$ $$\iff a^2+ab+b^2\mid \gcd(a,b)^3$$
Thus, let $d:=\gcd(a,b)$, and let $x:=\frac{a}d, y:=\frac{b}d$. It follows that $$a^2+ab+b^2\mid \gcd(a,b)^3\iff x^2+xy+y^2\mid d\implies d>xy$$
You are left to show that $\lvert a-b\rvert>\sqrt[3]{ab}\iff \lvert dx-dy\rvert^3=\lvert a-b\rvert^3> ab = d^2\cdot xy\iff d\lvert x-y\rvert > xy$. But this is trivial, since $d>xy$ and $\lvert x-y\rvert\geqslant 1.$
