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.



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 $a=b$.

  • $\begingroup$ got it...thank you very much... $\endgroup$ – Abhijit A J Apr 24 '16 at 9:11
  • $\begingroup$ You welcome, glad I could help you. $\endgroup$ – user261263 Apr 24 '16 at 9:24

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