Solve the following system of equations for real numbers a, b, c, d, e. $$ 3a = (b + c + d)^3,\\ 3b = (c + d + e)^3,\\ 3c = (d + e + a)^3,\\ 3d = (e + a + b)^3,\\ 3e = (a + b + c)^3. $$
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2Please look up Mathjax/Latex, and format your question properly next time. You are more likely to get answers if you show what you have tried, and what are you exactly struggling with – Andrei Nov 08 '17 at 15:16
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Do you want a solution, or do you want all solutions? Because $a=b=c=d=e \in {0, 1/3, -1/3}$ work as solutions. – Craig Nov 08 '17 at 15:38
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All solutions are i want to find – Shubham Patel Nov 08 '17 at 15:40
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If not other solutions possible then want toprove so – Shubham Patel Nov 08 '17 at 15:42
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You can prove that $a = b = c = d=e$ by noting $3a < 3b \implies b+c +d < c+d +e \imples b < e \implies c+d+e < a+b +c$ ... etc. will lead to a contradiction. – fleablood Nov 08 '17 at 22:40
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No wonder the problem looked familiar. This is a duplicate of this question. – Tito Piezas III Nov 09 '17 at 13:40
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Notice the map $x \mapsto x^3$ is strictly increasing. In other words, $x > y \iff x^3 > y^3$.
Consider what happens when $a > b$, we have
$$ \begin{array}{cll} a > b &\implies (b+c+d)^3 > (c+d+e)^3 &\implies b > e\\ \large\Downarrow\\ a > b > e & \stackrel{a > e}{\implies} (b+c+d)^3 > (a+b+c)^3 & \implies d > a\\ \large\Downarrow\\ d > a > b > e & \stackrel{d > e}{\implies} (e + a + b)^3 > (a + b + c)^3 & \implies e > c\\ \large\Downarrow\\ d > a > b > e > c & \stackrel{b > c}{\implies} (c + d + e)^3 > (d + e + a)^3 & \implies c > a\\ \large\Downarrow\\ d > a > b > e > c > a \end{array} $$ This leads to a contradiction that $a > a$. By a similar argument, if $a < b$, we find another contradiction $a < a$. As a result, any solution of the set of equations must have $a = b$.
Notice the set of equations is invariant under cyclic permutation $a \to b \to c \to d \to e \to a$.
This means any solution of the set of equations must satisfy $a = b = c = d = e$. Let $\lambda$ be this common value. The set of $5$ equations reduces to
$$3\lambda = (\lambda + \lambda + \lambda)^3 = 27\lambda^3 \quad\iff\quad 3\lambda(1 - 9\lambda^2) = 0 \quad\iff\quad \lambda = 0\text{ or } \pm \frac13$$
The given set of equations have $3$ and only $3$ real solutions:
$$a = b = c = d = e = 0\text{ or } \pm\frac13$$
achille hui
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Wolog $a \le b \le c \le d \le e$. Suppose $a < e$
Then $3a < 3e$ so $(b+c+d)^3 < (a+b+c)^3$ so $b+c +d < a+b+c$ and $d < a$ which is a contradiction of $a \le b \le c \le d \le e$. So $a = e$ and $a \le b \le c \le d\le e = a$ implies:
$a= b = c = d = e$.
And $3a = (3a)^3$ so $27a^3 - 3a = 3a(3a+1)(3a -1) = 0$ so $a = 0$ or $a =\pm \frac 13$
fleablood
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The system of equation is symmetric for a, b, c, d and e; If we suppose:
$a=b=c=d=e$ , we have:
$3a =(3a)^3 ⇒ 27 a^3 = 3 a ⇒ a^2 =1/9 ⇒ a = ± 1/3$ or:
$ a=b=c=d=e = ± 1/3$
So that for each equation we have:
$ 3 ({ ± 1/3}) = ( ± 1/3 ± 1/3 ± 1/3)^3 = ± 1$
sirous
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1You can't conclude that the variables are equal. Some solutions have that property, but not necessarily all. Here it turns out to be true. – Ross Millikan Nov 08 '17 at 16:06
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Symetric wouldn't mean they are equal. It would mean if $a=x$ is a solution than $b=x$ would also be a soultion. – fleablood Nov 08 '17 at 22:06