How many triplets of real numbers $(x, y, z)$ which satisfy these $3$ restriction: How many  triplets of real numbers $(x, y, z)$ which satisfy :
$$(x + y)^3 = z$$
$$(y + z)^3 = x$$
$$(z + x)^3 = y$$
I need some approaches for solving this problem. 
 A: Look first for solutions  of the shape $x=y=z=a$. Our equations all reduce to
$(2a)^3=a$, which has the solutions $a=0$ and $a=\pm 2^{-3/2}$.
Now look for solutions where not all the variables are equal.  For definiteness, look for solutions with  $x \lt z$.  
If $x \lt z$, then $x+y \lt y+z$ and therefore $(x+y)^3 \lt (y+z)^3$. From the first two equations, it follows that $z \lt x$, which is impossible.
Thus there are $3$ triples that satisfy the system of equations.
One can write up the same idea by starting from $x \le z$ and concluding that $z\le x$, which shows that $x=z$.  So for a solution, we must have $x=y=z$. 
Other approaches: We sketch a more "algebraic" approach which happens to be more work.  From the first two equations, we obtain
$$(x+y)^3-(y+z)^3=z-x.$$
Let $Z=x+y$ and $X=y+z$. Factoring the expression $Z^3-X^3$ on the left, we obtain
$$(Z-X)(Z^2+ZX+X^2)= (x-z)(Z^2+ZX+X^2)=z-x.$$
If $z \ne x$, this forces $Z^2+ZX+X^2=-1$. But this last equation does not have real solutions. There are many ways to see this, such as the Quadratic Formula. A cuter way is to note that 
$$4(Z^2 +ZX+X^2)=(2Z+X)^2+3X^2 \ge 0.$$
So we must have $z=x$. Similarly, $z=y$, and we end up looking for solutions of $(2a)^3=a$.
