Values of algebraic expressions

Algebra Level pending

x 3 + y 3 + 15 x y − 125 x^3 + y^3 + 15xy -125

What is the value of the two-variable expression above, if x + y = 5 x+y=5 ?


The answer is 0.

Sahar Bano by Sahar Bano , 38, India

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3 solutions

Chew-Seong Cheong
Mar 11, 2020

x 3 + y 3 + 15 x y − 125 = ( x + y ) ( x 2 − x y + y 2 ) + 15 x y − 125 = ( x + y ) ( ( x + y ) 2 − 3 x y ) + 15 x y − 125 = ( x + y ) 3 − 3 ( x + y ) x y + 15 x y − 125 Note that x + y = 5 = 5 3 − 3 ( 5 ) x y + 15 x y − 125 = 125 − 15 x y + 15 x y − 125 = 0 \begin{aligned} x^3 + y^3 + 15xy - 125 & = (x+y)(x^2-xy+y^2) + 15xy - 125 \\ & = (x+y)((x+y)^2 - 3xy) + 15xy - 125 \\ & = \blue{(x+y)}^3 - 3\blue{(x+y)}xy + 15xy - 125 & \small \blue{\text{Note that }x+y = 5} \\ & = \blue 5^3 - 3(\blue 5) xy + 15xy - 125 \\ & = 125 - 15xy + 15xy - 125 \\ & = \boxed 0 \end{aligned}

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Patrick Corn
Mar 12, 2020

The expression factors: x 3 + y 3 + 15 x y − 125 = ( x + y − 5 ) ( x 2 − x y + y 2 + 5 x + 5 y + 25 ) , x^3+y^3+15xy-125 = (x+y-5)(x^2 - xy + y^2 + 5x + 5y + 25), so the answer is 0 . \fbox{0}.

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Sahar Bano
Mar 11, 2020

x^3 +y^3 +15xy-125

As (x+y)(x^2-2xy+y^2)=x^3 +y^3

Therefore we can simplify the expression as: (x+y)(x^2-2xy+y^2)+15xy-125

5(x^2-xy+y^2)+15xy-125. (Because x+y=5)

5x^2-5xy+5y^2+15xy-125

5x^2+5y^2+10xy-125

5(x^2+y^2+2xy)-125

5(x+y)^2 -125

5(5^2) -125. (because x+y=5)

125-125=0

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