<s>[INST] 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?
 [/INST] So first you factorize the expression:
$$(x-y)(x^2+xy+y^2-3x-3y)=0$$
$$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$
When the product of two expressions are $0$ either of them is $0$
So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0.
Second factor can be treated as a quadratic equation in terms of either $x$ or $y$,
$$x^2+x(y-3)+(y^2-3y)=0$$
Where you consider each $y$ as a constant and solve the equation to get:
$$x=\frac{-(y-3)\pm\sqrt{(y-3)^2-4(y^2-3y)}}{2}$$
$$x=\frac{-(y-3)\pm\sqrt{3(9-y^2)}}{2}$$ 
For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,\pm1,\pm2,\pm3$  but you also want it to be a perfect square so only $y=\pm3$ which gives:
$$x = 0,3 $$ 
Final solutions:
All integers for which $x=y$,
$(0,3)$,
$(3,-3)$
You can also solve the equation for $y$, hope this helps .....
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