Find all real solutions for the system: $x^3=y+y^5$, $y^5=z+z^7$, $z^7=x+x^3.$ 
Given: $$\left\{
\begin{array}{l}
x^3=y+y^5\\ 
y^5=z+z^7\\
z^7=x+x^3 
\end{array}
\right. 
$$
  Find: all real solutions for the system.

From a book on preparation for math contests. The answer states there is just one solution. My problem is showing that this is indeed the case.
My attempt: it is easy to see that one solution is $(x,y,z)=(0,0,0)$. And, adding the equations, we get to $x+y+z=0$. But my problem is showing that this solution is indeed unique, if the answer provided in the book is right. 
Hints and answers are appreciated. Sorry if this is a duplicate. 

Multiplying the three equations gives $x^3 y^5 z^7 = x y z (1+x^2)(1+y^4)(1+z^6)\,$.
If $xyz \ne 0$ then it follows that $x^2 y^4 z^6 = (1+x^2)(1+y^4)(1+z^6)\,$, but the latter is not possible since $0 \le x^2 \lt 1+ x^2, 0 \le y^4 \lt 1+y^4, 0 \le z^6 \lt 1+ z^6\,$.
Therefore $xyz=0\,$, and it is easy to show that any one of $x,y,z$ being $0$ implies all of them being $0$.
