Find the real values of $a,b,c,d,e$ where $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, $3c=(d+e+a)^3$, $3d=(e+a+b)^3$, $3e=(a+b+c)^3$. It is the problem. I have no idea about that. I tried to do this using basic algebra. I tried to use the methods of polynomials but I failed. Somebody help me.
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Due to cyclicity of the system we can assume that $a\ge b,c,d,e$. We can get all other solutions by cyclic permutations.
We have $$3a = (b+c+d)^3 \le (b+c+a)^3 = 3e \le 3a$$ so we must have equalities here. Therefore $a=e$.
Repeating the argument we get consequently $e=d=c=b$. Therefore all numbers are equal. Then you are left with solving equation $3a=(3a)^3$ which gives $a\in\{-\frac 13, 0, \frac 13\}$.
Therefore the system has three solutions: $$(a,b,c,d,e) \in \left\{ \left(-\frac 13,-\frac 13,-\frac 13,-\frac 13,-\frac 13\right), (0,0,0,0,0), \left(\frac 13,\frac 13,\frac 13,\frac 13,\frac 13\right) \right\}$$
timon92
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In the first step, are we assuming WLOG that since they are real numbers, there must be an order and hence $a$ is not less than any element? – ShankRam May 07 '17 at 15:31
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@ShankRam we use the obvious fact that any finite subset of $\mathbb R$ has an element greater or equal to every other element of that finite set. – timon92 May 08 '17 at 16:22
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Yeah, that's what I asked. Thanks :) – ShankRam May 08 '17 at 17:32
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One approach, prompted by the symmetry of the equations, is to assume all the variables are equal. This gives $3a=(3a)^3$, which has solutions $a=0,a=\pm3^{-1}$. I don't see an easy way to look for others and couldn't get Alpha to solve them.
Tito Piezas III
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Ross Millikan
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I am not satisfied about your answer. Why $a=b=c=d=e$?? Ot may have other solutions. – Sufaid Saleel Oct 10 '16 at 14:35
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@SufaidSaleel: I said it was an assumption. The fact that it works out justifies that I have found three solutions. You are correct that there may be more. timon92 makes a good argument that there are not – Ross Millikan Oct 10 '16 at 14:55