Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$. 
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.

I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?

$10 \mid (a^2+ab+b^2)$ implies $10 \mid a^3-b^3 = (a-b)(a^2+ab+b^2)$, that is, $a^3 \equiv b^3 \bmod 10$.
Now, $x \mapsto x^3$ is a bijection mod $10$. Therefore, $a^3 \equiv b^3 \bmod 10$ implies $a \equiv b \bmod 10$.
Then $a^2+ab+b^2 \equiv 3a^2$ implies $10 \mid a$ and so $10 \mid b$.
Therefore, $1000 \mid a^3$ and $1000 \mid b^3$, hence $1000 \mid (a^3-b^3)$.
