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Question Number 9523 by Joel575 last updated on 12/Dec/16

$$3a={(b+c+d)}^{2014}$$$$3b={(a+c+d)}^{2014}$$$$3c={(a+b+d)}^{2014}$$$$3d={(a+b+c)}^{2014}$$$$\mathrm{Find}\mathrm{all}\mathrm{the}\mathrm{solution}\mathrm{of}(a,b,c,d)\mathrm{if}a,b,c,d\in \mathbb{R}$$

Answered by mrW last updated on 16/Dec/16

$$\mathrm{let}\mathrm{u}=\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}$$$$\Rightarrow 3\mathrm{a}={(\mathrm{u}-\mathrm{a})}^{2014}$$$$\Rightarrow 3\mathrm{b}={(\mathrm{u}-\mathrm{b})}^{2014}$$$$\Rightarrow 3\mathrm{c}={(\mathrm{u}-\mathrm{c})}^{2014}$$$$\Rightarrow 3\mathrm{d}={(\mathrm{u}-\mathrm{d})}^{2014}$$$$\Rightarrow \mathrm{a}=\mathrm{b}=\mathrm{c}=\mathrm{d}=\mathrm{x}$$$$3\mathrm{x}={\left(3\mathrm{x}\right)}^{2014}$$$$\Rightarrow \mathrm{x}=0$$$$\Rightarrow {\left(3\mathrm{x}\right)}^{2013}=1$$$$\Rightarrow 3\mathrm{x}={}^{2013}\sqrt{1}=1$$$$\Rightarrow \mathrm{x}=\frac{1}{3}$$

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