The answer isn't 3!

Algebra Level 2

If x + y = 1 x+y=1 and x 2 + y 2 = 2 x^{2}+y^{2}=2 , find the value of x 3 + y 3 x^{3}+y^{3} .


The answer is 2.5.

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2 solutions

Zach Abueg
Feb 14, 2017

x + y = 1 \displaystyle {\color{#D61F06}{x + y}} = \color{#D61F06}{1}

x 2 + y 2 = 2 \displaystyle {\color{#20A900}{x^2 + y^2}} = \color{#20A900}{2}

x 2 + y 2 = ( x + y ) 2 2 x y 1 2 2 x y = 2 x y = 1 2 \displaystyle x^2 + y^2 = (x + y)^2 - 2xy \Longrightarrow {\color{#D61F06}{1}}^2 - 2xy = 2 \Longrightarrow {\color{#3D99F6}{xy}} = {\color{#3D99F6}{-\frac 12}}

x 3 + y 3 = ( x + y ) ( x 2 x y + y 2 ) = ( x + y ) ( x 2 + y 2 x y ) = ( 1 ) ( 2 1 2 ) = 2.5 \displaystyle x^3 + y^3 = (x + y)(x^2 - xy + y^2) = (x + y)(x^2 + y^2 - xy) = ({\color{#D61F06}{1}})({\color{#20A900}{2}} - {\color{#3D99F6}{-\frac12}}) = 2.5

x 3 + y 3 = 2.5 \displaystyle x^3 + y^3 = \boxed{2.5}

If x 3 + y 3 = a , x 2 + y 2 = b , x + y = c x^3+y^3=a \; , x^2+y^2=b \;, x+y=c then the identity a = 1 2 ( 3 b c c 3 ) \displaystyle a=\frac{1}{2}(3bc-c^3) holds. Substituting c = 1 , b = 2 c=1,b=2 we have the answer 2.5 \boxed{2.5}

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