Find all possible real solutions. Find all possible real solutions of $a, b, c, d$ and $e$ if:
$3a= (b+c+d)^3$
$3b= (c+d+e)^3$
$3c= (d+e+a)^3$
$3d= (e+a+b)^3$
$3e= (a+b+c)^3$
Well I believe the solutions are possible only if $a=b=c=d=e$. In that case the solutions possible are $0, \frac{1}3$ and $-\frac{1}3$, but I am unable to prove that no other solution exists. Probably using $A.M.\geq G.M.$ might work (where equality is possible iff the terms are equal). But again I am not able to get that to help.

HINT
If $x, y$ real then $x^3 < y^3 \Leftrightarrow x<y$.
I consider the equalities to be numbered.
Now suppose $a>b$. 
Then from equality (1) and (2) we get $b>e$.
Because $a>e$ from (1) and (5) we get $d>a$.
So $d > a > b > e$.
Now, because $d>e$ from (4) and (5) we get $e > c$, so $d > a > b > e > c$.
Because $b > c$ from (2) and (3) we get $c>a$ which is false.
Therefore $a=b$.
