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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) Find all the solution of (a, b, c, d) if a, b, c, d ∈ R

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

Answered by mrW last updated on 16/Dec/16

let u=a+b+c+d ⇒3a=(u−a)^(2014) ⇒3b=(u−b)^(2014) ⇒3c=(u−c)^(2014) ⇒3d=(u−d)^(2014) ⇒a=b=c=d=x 3x=(3x)^(2014) ⇒x=0 ⇒(3x)^(2013) =1 ⇒3x=^(2013) (√1)=1 ⇒x=(1/3)

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

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