TITLE: Cyclic system of cubic equations in $5$ variables
QUESTION [9 upvotes]: Problem: To find all real solutions of the system:
$$3a=(b+c+d)^3$$
$$3b=(c+d+e)^3$$
$$3c=(d+e+a)^3$$
$$3d=(e+a+b)^3$$
$$3e=(a+b+c)^3$$
My attempt: I tried to get a bound for positive solutions. Using AM$\geq$GM, $(x+y+z)^3\geq27xyz$, I get 
$$ abcde\leq \sqrt{\frac{1}{9^5}}=\frac{1}{3^5}$$ 
Also, by Jensen's inequality $\frac{x^3+y^3+z^3+u^3+v^3}{5}\geq(\frac{x+y+z+u+v}{5})^3$ I get
$$a+b+c+d+e\leq5/3$$
Other than this, I cannot find any more information regarding the roots. This problem is not supposed to be a very tough one, so probably I am missing something here...

REPLY [2 votes]: A real solution must satisfy $a=b=c=d=e$. To see this, let an arbitrary solution $(a,b,c,d,e)$ be given, and replacing it with a cyclic shift, we can assume without loss of generality that $a=\max\{a,b,c,d,e\}$. Then $a\geq e$, so that $\sqrt[3]{3a} \geq \sqrt[3]{3e}$. Using the equations $3a=(b+c+d)^3$ and $3e=(a+b+c)^3$, this implies that $b+c+d \geq a+b+c$, i.e., $d \geq a$. But since $a=\max\{a,b,c,d,e\} \geq d$, this implies $d=a$. Iterating this same chain of reasoning with $d$ in place of $a$, we obtain, in turn, $b=d$, $e=b$, and $c=e$, so that indeed $a=b=c=d=e$. 
It follows that the only real solutions satisfy $3a=(b+c+d)^3=(3a)^3$, so that $9a^3=a$, which has only solutions $a\in\{0,\pm \frac13\}$. Therefore, the real solutions are $(0,0,0,0,0)$, $(\frac13,\frac13,\frac13,\frac13,\frac13)$, and $(-\frac13,-\frac13,-\frac13,-\frac13,-\frac13)$.