Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.

So far I've got one pair of solution. Try  $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then 
$$ x^3 - x^3 = 3(x^2 -x^2) \\
0 = 0 \qquad \text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$\begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\
x^2+xy+y^2 &= 3(x+y) \end{aligned}\\
\begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \\
x^2+x(y-3)+y(y-3) &=0 \\
 x^2 + (y-3)(x+y) &= 0 \end{aligned}$$
What now? Am I on the right track?

By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=\frac{x^2+xy+y^2}{3}>0$
Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.
Hope it is helpful:)
