<s>[INST] Solving $1 - 3xy + x^3 + y^3 = 0$ I am trying to solve the equation
$$
1 - 3xy + x^3 + y^3 = 0
$$
in the real numbers. I can see several (infinitely many) solutions:


*

*$x=y = 1$

*$x+ y = -1$
I am trying to show that these are all of them. 
For example, if $x = y$, then $x^2(-3+2x) = -1$ forcing $x = y = 1$
I can't seem to make much progress with the case where $x\neq y$ and showing that $x + y = 1$ in that case. (I am pretty sure that this must be true.)
 [/INST] If you substitute $y$ with $-x-1$ you get the identity
$$x^3+(-x-1)^3-3x(-x-1)+1=0$$
This means that the polynomial $x^3+y^3-3xy+1$ is divisible by $x+y+1$. Performing long division you get the following factorization:
$$x^3+y^3-3xy+1= (x+y+1)(x^2-xy+y^2-x-y+1)$$
so the problem becomes proving that $(x;y)=(1;1)$ is the only zero for
$$x^2-xy+y^2-x-y+1=0$$
Call $X=x-1$ and $Y=y-1$. We want to prove that $X=Y=0$. Substituting $x=X+1$ and $y=Y+1$, we get the equation
$$X^2-XY+Y^2=0$$
whose unique solution is $X=Y=0$ (hope you can find the reason of this by yourself). This concludes the proof.
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