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?

Here's a more general result:

Let $\Psi_n(x,y)$ denote the $n$th homogenized cyclotomic polynomial. If $x,y\in\mathbb Z$, every prime divisor of $\Psi_n(x,y)$ is either:
  
  
*
  
*$1\pmod n$
  
*a prime divisor of $n$
  
*a prime divisor of both $x$ and $y$
  

This is a simple corollary of the same result for the usual cyclotomic polynomial $\Phi_n(x)=\Psi_n(x,1)$ (where the third case doesn't occur).
In our case $2\mid\Psi_3(a,b)=a^2+ab+b^2$. As $3\nmid2$ and $2\not\equiv1\pmod3$, we have $2\mid a,b$. Same argument for $5$.

Entirely self-contained proof of that result in the case of prime $n$ (here $n=3$):
We then have $\Psi_n(x,y)=x^{n-1}+xy^{n-2}+\cdots+y^{n-1}$. Let $p\mid\Psi_n(x,y)\mid x^n-y^n$. If $p\mid x$ or $y$, we are done. If not, $y$ is invertible mod $p$, and $(x/y)^n\equiv1$. Let $k$ be the order of $x/y$ mod $p$. $k\mid n$.
If $k=1$, $x\equiv y$ and $p\mid x^{n-1}+xy^{n-2}+\cdots+y^{n-1}\equiv nx^{n-1}$, so $p\mid n$, so $p=n$.
If $k=n$, $n\mid p-1$ by Fermat's Little Theorem.
