Question

$\frac{x^2-(y-z{)}^2}{(x+z{)}^2-y^2}+\frac{y^2-(x-z{)}^2}{(x+y{)}^2-z^2}+\frac{z^2-(x-y{)}^2}{(y+z{)}^2-x^2}=$___________

• A. -1
• B. $\frac{1}{2}$
• C. 0
• D. 1

Answer 1

The correct option is D 1

$\frac{x^2-(y-z{)}^2}{(x+z{)}^2-y^2}+\frac{y^2-(x-z{)}^2}{(x+y{)}^2-z^2}+\frac{z^2-(x-y{)}^2}{(y+z{)}^2-x^2}\left[\text{Using}\right(a^2-b^2)=(a+b\left)\right(a-b\left)\right]$
$=\frac{[x-(y-z\left)\right](x+y-z)}{(x+z-y)(x+z+y)}+\frac{[y-(x-z\left)\right](y+x-z)}{(x+y-z)(x+y+z)}+\frac{[z-(x-y\left)\right](z+x-y)}{(y+z-x)(y+z+x)}$
$=\frac{(x-y+z)(x+y-z)}{(x+z-y)(x+z+y)}+\frac{(y-x+z)(y+x-z)}{(x+y-z)(x+y+z)}+\frac{(z-x+y)(z+x-y)}{(y+z-x)(y+z+x)}$
$=\frac{x+y-z}{x+y+z}+\frac{y+z-x}{x+y+z}+\frac{z+x-y}{x+y+z}$
$=\frac{x+y-z+y+z-x+z+x-y}{x+y+z}$
$=\frac{x+y+z}{x+y+z}$
$=1$