Prove that the only integer solutions to $x^3 + 3x^2y - 3xy^2 - 3y^3 = 1$ are $(1,0)$ and $(-2,3)$.
My only idea for now is to try to represent this in the form $A^3 + B^3 = C^3$ and apply Fermat's Last Theorem. However, I cannot find suitable $A,B,C$.
It is also easy to see that firstly $x\equiv 1 \pmod 3$ and then $3$ divides $y$, but substituting does not give a better looking equation.
Another alternative: if we set $s = x+y$, then we get $s^3 -6sy^2 + 2y^3 = 1$ which looks a bit better.
Any help appreciated!