Solve a system of equations for real $x,y,z$ Solve the following system of equations in $x,y,z \in \Bbb R$:
$$ (x+y)^3 = z $$
$$ (x+z)^3 = y$$
$$ (y+z)^3 = x$$
I found the solutions $(0,0,0)$, $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ and $(-\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}})$, but I'm not sure if my proof is correct. Also I was wondering if there is a more elegant solution than the one I found (described below).
First, I took the first two equations and solved for $x$, getting
$$ x = \sqrt[3]{z} - y$$
$$ x = \sqrt[3]{y} - z$$
Then substituting $z = c^3$ and $y = b^3$ (since we are working in $\Bbb R$, these values of $b$ and $c$ exist and are unique), we get 
$$c - b^3 = b - c^3$$
and thus 
$$c + c^3 = b + b^3$$
Since the function $f(x) = x^3 + x$ is one-to-one and onto, this implies $b=c$. By symmetry, this means $a=b=c$. Since $f(x) = x^3$ is one-to-one and onto, this implies $x=y=z$. Plugging this into the original equation gives
$$8x^3 = x$$
which has the solutions $x = 0, \pm \frac{1}{2\sqrt{2}}$. Hence the three solution pairs I found.
Is this solution ok? Is there an alternate, more elegant approach (that perhap more obviously uses the symmetry between $x$, $y$ and $z$?)

This problem is very pretty:
Let's suppose that $x\geq y\geq z$ (any order will let you the same, you can check that). Because of this, 
$$x+y\geq x+z\geq z+y$$
$$\Rightarrow (x+y)^3\geq (x+z)^3\geq (z+y)^3$$
But that means that:
$$z\geq y\geq x$$
So from that $x=y=z$ then for any equation you will have $8x^3=x$. From this last equation you have $x=0$ is solution or $x=\pm \frac{1}{2\sqrt{2}}$
