Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get 
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get 
$$a\equiv b \text{ mod } 3  \text{ or } a^2+ab+b^2\equiv0 \text{ mod } 3.$$
Besides, the prime factorization of $999=3^3\cdot37$,  but I don't know how to go on. 
I would appreciate any hints.

$$999=a^3-b^3=(a-b)^3+3ab(a-b)\geq(a-b)^3,$$ which says $a-b$ is divided by $3$ and $a-b\leq9$ and we get not so many cases:
$a-b=3$ or $a-b=9$.
