Algebra problem with divisibility

Question

Let $a,b$ be natural numbers so that $a \ge 2, b \ge 2$ and $a^{2}b^{2}$ is divisible by $a^{2}-ab+b^{2}$ Prove that $a$, $b$ are not relatively prime numbers. Any ideas?

Answer 1

${\bf Hint}\quad \begin{align} &(\color{#c00}a,\!\overbrace{a^2\!-\!ab\!+\!b^2}^{\Large\ \equiv\ \color{#0a0}{ b^2}\bmod{\color{#c00} a}}\!\!)=(a,\color{#0a0}{b^2})\\ &(b,\ a^2\!-\!ab\!+\!b^2)=(b,a^2) \end{align}\ $ $ \begin{align}\\ \\ {\rm so}\ \ \ (a^2b^2,\,a^2\!-\!ab\!+\!b^2)=1 \iff (a,b)=1\\ \\ \end{align}$