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Algebra Level 4

x 2 + y 2 + z 2 + 2004 w 2 = 4 w ( x + y + z ) x^{ 2 }+y^{ 2 }+z^{ 2 }+2004w^{ 2 }=4w(x+y+z)

determine ( x + 20 ) ( y + 8 ) ( z + 9 ) ( w + 7 ) (x+20)(y+8)(z+9)(w+7)


The answer is 10080.

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Watcharapon Wisadekul by Watcharapon Wisadekul , 28, Thailand

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2 solutions

Rajen Kapur
Nov 11, 2014

Given that (x - 2w)^2 + (y - 2w)^2 + (z - 2w)^2 + 1992 w^2 = 0, it is evident that x = y = z = 2w and w = 0.

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Thushar Mn
Nov 10, 2015

The first given equation satisfies x=y=z=w=0. so put x=y=z=w=0 in the expression 2 b found out... :)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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