<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] Worth memorizing:
$$ x^3 + y^3 + z^3 - 3xyz = (x+y+z) \left(x^2 + y^2 + z^2 - yz - zx - xy  \right)  $$
Another step is possible, in that
$$ x^2 + y^2 + z^2 - yz - zx - xy = \frac{1}{2} \; \left((x-y)^2 + ( y-z)^2 + ( z-x)^2 \right)  $$
The cubic is zero, for real numbers $x,y,z,$  when either
$x+y+z = 0$  or $x=y=z.$ In your case, take $z=1$
More solutions are possible over the complexes, in that the quadratic further factors as
$$ (x+\omega y + \omega^2 z)(x + \omega^2 y + \omega z) $$
where $\omega$ is a primitive cube root of unity. 
</s>