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[Community Question] Algebra-Precalculus: Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $

One of our user asked:

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?


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