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.)

$$x=u+v$$
$$x^3=u^3+v^3+3uv(x)\iff x^3-3uv(x)-(u^3+v^3)=0$$
Comparing with $$x^3-3xy+y^3+1=0$$
$$uv=y$$
and $$ u^3+v^3=-(y^3+1)\iff u^3+\dfrac{y^3}{u^3}+y^3+1=0$$
$$0=u^6+(y^3+1)u^3+y^3=(u^3+1)(u^3+y^3)$$
If $u^3+1=0,u=-1,\implies v=-y$
$$\implies x=u+v=-1-y\iff x+y=-1$$
$$x^3+y^3-3xy+1=(x+y)^3-3xy(x+y+1)+1=0$$
