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Algebra Level 1

x 2 − ( y − z ) 2 ( x + z ) 2 − y 2 + y 2 − ( x − z ) 2 ( x + y ) 2 − z 2 + z 2 − ( x − y ) 2 ( y + z ) 2 − x 2 = ? \dfrac{x^2- (y-z)^2}{(x+z)^2- y^2} + \dfrac{y^2- (x-z)^2}{(x+y)^2- z^2} + \dfrac{z^2- (x-y)^2}{(y+z)^2- x^2} = ?


The answer is 1.

Jason Chrysoprase by Jason Chrysoprase , 18, Indonesia

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2 solutions

Rishabh Jain
Jan 22, 2016

∑ c y c x 2 − ( y − z ) 2 ( x + z ) 2 − y 2 \displaystyle \sum_{cyc} \frac{x^2- (y-z)^2}{(x+z)^2- y^2} = ∑ c y c ( x − y + z ) ( x + y − z ) ( x − y + z ) ( x + y + z ) =\displaystyle \sum_{cyc} \dfrac{\color{#D61F06}{(x-y+z)}(x+y-z)}{\color{#D61F06}{(x-y+z)}(x+y+z)} = ∑ c y c x + y − z x + y + z =\displaystyle \sum_{cyc} \dfrac{x+y-z}{x+y+z} = 1 \quad~\Large=1

6 Helpful 0 Interesting 0 Brilliant 0 Confused
Karthick Shiva
Jan 23, 2016

I just substituted x=1, y=1, z=0 and got the answer. :)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

For these values, first term's denominator is turning to 0 !!!

Rishabh Jain - 5 years, 4 months ago

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