When $\frac {a^3-b^3}{a^2-b^2}$ is an integer? If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer. 


  For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?
Assuming $a\neq b$, we want that $\frac{a^2+ab+b^2}{a+b}=(a+b)-\frac{ab}{a+b}$ is an integer. The solutions of
$$ \frac{ab}{a+b} = k, $$
assuming $a+b\neq 0$, are the solutions of 
$$ ab-ka-kb = 0, $$
i.e. the solutions of 
$$ (a-k)(b-k) = k^2, $$
which depend on the couples of divisor/complementary divisor of $k^2$. In general, for any $d\mid k^2$ we have the solution $$a=d+k,\qquad b=\frac{k^2}{d}+k.$$
A trivial example is for $a,b \in \mathbb{Z}$ such that $a+b=1$