solve the system of equation for $a,b,c,d,e$

Question

Solve the system of equation for $a,b,c,d,e$

$$3a=(b+c+d)^3$$ $$3b=(c+d+e)^3$$ $$3c=(d+e+a)^3$$ $$3d=(e+a+b)^3$$ $$3e=(a+b+c)^3$$ I try this problem

I find this system of equation is symmetric

i.e, if (a,b,c,d,e) satisfy this system , then it's permutation is also satisfy this equation

I want a problem solving tip to start this problem

Answer 1

They cannot have different values.

• Suppose for example $a > c$:

  • then $(b+c+d)^3 > (d+e+a)^3$ so $b+c+d>d+e+a$ and $b-e>a-c>0$ requiring $b>e$;
  • but that needs $(c+d+e)^3 > (a+b+c)^3$ requiring $d > a$;
  • and continuing $e >c$
  • then $b>d$
  • then $c>a$ contrary to the original supposition

• Similarly $c <a$ would lead to $a < c$, reversing the inequalities in the previous point

So $a=c$ and applying this cyclically $a=c=e=b=d$.

We are reduced to solving $3a=(3a)^3$, i.e. $(3a)(3a-1)(3a+1)=0$, giving the solutions $$a=b=c=d=e=0 \text{ or } \frac13\text{ or }-\frac13$$ These three sets of solutions all work with the original equations