Square numbers Problem

Algebra Level 3

Real numbers a a , b b , and c c are such that a 2 + b 2 + c 2 = 1 a^{2}+b^{2}+c^{2}=1 . Find the value of the expression below:

( a + b + c ) 2 + ( − a + b + c ) 2 + ( a − b + c ) 2 + ( a + b − c ) 2 (a+b+c)^{2}+(-a+b+c)^{2}+(a-b+c)^{2}+(a+b-c)^{2}


The answer is 4.

Tin Le by Tin Le , 15, Vietnam

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1 solution

Chew-Seong Cheong
Jun 21, 2019

X = ( a + b + c ) 2 + ( − a + b + c ) 2 + ( a − b + c ) 2 + ( a + b − c ) 2 = a 2 + b 2 + c 2 + 2 ( a b + b c + c a ) + a 2 + b 2 + c 2 + 2 ( − a b + b c − c a ) + a 2 + b 2 + c 2 + 2 ( − a b − b c + c a ) + a 2 + b 2 + c 2 + 2 ( a b − b c − c a ) = 4 ( a 2 + b 2 + c 2 ) Since a 2 + b 2 + c 2 = 1 = 4 \begin{aligned} X = & \ (a+b+c)^2 + (-a+b+c)^2 + (a-b+c)^2 + (a+b-c)^2 \\ = & \ a^2 + b^2 + c^2 + 2(ab+bc+ca) + \\ & \ a^2 + b^2 + c^2 + 2(-ab+bc-ca) + \\ & \ a^2 + b^2 + c^2 + 2(-ab-bc+ca) + \\ & \ a^2 + b^2 + c^2 + 2(ab-bc-ca) \\ = & \ 4 ({\color{#3D99F6} a^2 + b^2 + c^2}) & \small \color{#3D99F6} \text{Since }a^2 + b^2 + c^2=1 \\ = & \ \boxed 4 \end{aligned}

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