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.)
 A: First, consider the transformation
$$x=u+v$$
$$y=u-v.$$
Then $1-3x_0y_0+x_0^3+y_0^3=0$ if and only if $(1 + 2 u_0) ((-1+ u_0)^2 + 3 v_0^2)=0$ where
$$u_0=\frac{x_0+y_0}{2}$$
$$v_0=\frac{x_0-y_0}{2}.$$
Now, one possible solution is
$$1+2u=0\Rightarrow \frac{x+y}{2}=u=\frac{-1}{2}\Rightarrow x+y=-1$$
which is what you found. In fact, there is only one other solution as 
$$(-1 + u)^2 + 3 v^2\geq 0$$
with equality happening at $(u,v)=(1,0)$. Thus, the only other zero is
$$(u,v)=(1,0)\Rightarrow (x,y)=(1,1)$$
which is what you also found. We conclude the only zeros are at
$$x=y=1$$
$$x+y=-1.$$
A: For the case $x,y\ge 0$ by AM-GM
$$x^3+y^3+1\ge3\sqrt[3]{x^3y^3}=3xy$$
with equality if and only if $x=y=1$.
A: 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.
A: 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. 
A: You will get infinite solution when $x$ and $y$ are not equal as the relation $x+y=-1$ gives different values of $x$, when you are changing $y$ .Check this https://www.wolframalpha.com/input/?i=1-3xy%2Bx%5E3%2By%5E3%3D0
A: $$x=u+v$$
$$x^3=u^3+v^3+3uv(x)\iff x^3-3uv(x)-(u^3+v^3)=0$$
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 $u^3+1=0,u=-1,\implies v=-y$
$$\implies x=u+v=-1-y\iff x+y=-1$$
$$x^3+y^3-3xy+1=(x+y)^3-3xy(x+y+1)+1=0$$
