Question: Simplify: $\dfrac{(x-y)² -z²}{x² - (y+z)²} +$$\dfrac{(y-z)²-x²}{y² -(z+x)²} +$$\dfrac{(z-x)² - y²}{z²-(x+y)²}$


Solution:
Given,

$= \dfrac{(x-y)² -z²}{x² - (y+z)²} +\dfrac{(y-z)²-x²}{y² -(z+x)²} +\dfrac{(z-x)² - y²}{z²-(x+y)²}$

[ Using a²-b² = (a-b)(a+b) ]

$= \dfrac{(x-y-z)(x-y+z)}{(x-y-z)(x+y+z)} + \dfrac{(y-z-x)(y-z+x)}{(y-z-x)(y+z+x)} + \dfrac{(z-x-y)(z-x+y)}{(z-x-y)(z+x+y)}$

$= \dfrac{x-y+z}{x+y+z} + \dfrac{y-z+x}{x+y+z} + \dfrac{z-x+y}{z+x+y)}$

$= \dfrac{x-y+z}{x+y+z} + \dfrac{y-z+x}{x+y+z} + \dfrac{z-x+y}{x+y+z}$

$= \dfrac{x-y+z +y-z+x +z-x+y}{x+y+z}$

$= \dfrac{x+y+z}{x+y+z}$

$= 1$
= Answer