**Problem:** Find all possible real numbers \(a,b,c,d,e\) which satisfy the following set of equations:

\( \left\{ \begin{array}{ccc} 3a & = & ( b + c+ d)^3 \\ 3b & = & ( c + d +e ) ^3 \\ 3c & = & ( d + e +a )^3 \\ 3d & = & ( e + a +b )^3 \\ 3e &=& ( a + b +c)^3. \end{array}\right.\)

**Solution:** Observing the symmetry and the cyclicity of the given set of equations it can be easily inferred that the real numbers \(a,b,c,d\) and \(e\) cannot be ordered i.e. one number cannot be greater or smaller than the other number else the system of equations will be inconsistent. This can be shown easily with the help of inequality.

Without loss of generality, we can say that \(a\) is greater or equal to them all, i.e., \(a\geq b,c,d,e\). Thus we have

\((d+e+a) \geq ( c+d+e)\)

\(=> (d+e+a)^3 \geq (c+d+e)^3\)

\(=> 3c \geq 3b\)

\(=> c \geq b\)

As \(a \geq d\), we also have,

\((a+b+c) \geq ( b+c+d)\)

\(=> (a+b+c)^3 \geq (b+c+d)^3\)

\(=> 3e \geq 3a\)

\(=> e \geq a\)

Thus we have \(a=e\)

Further calculations will show that \(a=b=c=d=e\) is the only possible solution to this set of equations.

Thus we are left to solve just one equation, i.e., \(3a = (3a)^3\)

which gives \(a= -\frac{1}{3},0,\frac{1}{3}\)

Thus the possible values of the 5-tuple \((a,b,c,d,e)\) are:

\(( -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3}, -\frac{1}{3})\)

\((0,0,0,0,0)\)

\((\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})\)