Last problem before the Lunar New Year! - Minimalism is a decision - (7)

Algebra Level 3

{ x = a b c = a + b + c a b + b c + c a = 1 ⟹ a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) = □ \large \left \{ \begin{aligned} x &= abc = a + b + c\\ &ab + bc + ca = 1 \end{aligned} \right. \implies a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2) = \square

What is □ \square ?

2 x 2x − 2 x -2x − 2 -2 − x -x 0 0 − 1 -1 x x 1 1

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Atomsky Jahid
Feb 4, 2019

a ( b 2 + c 2 ) + b ( c 2 + a 2 ) + c ( a 2 + b 2 ) a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2) = a b ( a + b ) + b c ( b + c ) + c a ( c + a ) =ab(a+b)+bc(b+c)+ca(c+a) = a b ( x − c ) + b c ( x − a ) + c a ( x − b ) =ab(x-c)+bc(x-a)+ca(x-b) = x ( a b + b c + c a ) − 3 a b c =x(ab+bc+ca)-3abc = − 2 x =-2x

0 Helpful 1 Interesting 2 Brilliant 0 Confused

Yeah, it's that easy. Happy Lunar New Year!

Thành Đạt Lê - 2 years, 4 months ago

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