The once corrected.

Algebra Level 3

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

Which of the following options is □ \square ?

− 4 x -4x 4 x 4x 4 4 − 2 x -2x 2 x 2x − 4 -4 − 2 -2 2 2

Thành Đạt Lê by Thành Đạt Lê , 17, Singapore

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1 solution

Chew-Seong Cheong
Feb 11, 2019

□ = ∑ c y c a ( 1 − b 2 ) ( 1 − c 2 ) = ∑ c y c ( a − a b 2 − a c 2 + a b 2 c 2 ) = a + b + c − ( a b 2 + c 2 a + a 2 b + b c 2 + c a 2 + b 2 c ) + a b 2 c 2 + a 2 b c 2 + a 2 b 2 c = x − ( ( a b + b c + c a ) ( a + b + c ) − 3 a b c ) + a b c ( a b + b c + c a ) = x − ( x − 3 x ) + x = 4 x \begin{aligned} \square & = \sum_{cyc} a(1-b^2)(1-c^2) \\ & = \sum_{cyc} \left(a-ab^2-ac^2+ab^2c^2\right) \\ & = a+b+c - \left(ab^2+c^2a + a^2b + bc^2 + ca^2 + b^2c\right) + ab^2c^2 + a^2bc^2 + a^2b^2c \\ & = x - \left((ab+bc+ca)(a+b+c) -3abc \right) + abc(ab+bc+ca) \\ & = x - \left(x - 3x \right) + x \\ & = \boxed {4x} \end{aligned}

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Thành Đạt Lê , you can use \begin{cases} f(x) & \implies ...(1) \ g(x) & \implies ...(2) \end{cases} for { f ( x ) ⟹ . . . ( 1 ) g ( x ) ⟹ . . . ( 2 ) \begin{cases} f(x) & \implies ...(1) \\ g(x) & \implies ...(2) \end{cases} . I have done the changes for you. You can see the LaTex code by placing the mouse cursor on top of the formulas.

Chew-Seong Cheong - 2 years, 4 months ago

Well, I know that, but the format that I want to show for the problem is the "and" symbol (which is the enlarged version of {) to be flipped to the other side and put after the equations.

Thành Đạt Lê - 2 years, 4 months ago

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But I don't like your format.

Chew-Seong Cheong - 2 years, 4 months ago

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So what would you recommend? (You should rewrite this problem in your preferred format below.)

Thành Đạt Lê - 2 years, 4 months ago

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