Question

X/(x-y)(x-z)+y/(y-z)(y-x)+z/(z-x)(z-y)=0

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{\dfrac{x}{(x-y)(x-z)}+\dfrac{y}{(y-z)(y-x)}+\dfrac{z}{(z-x)(z-y)}=0}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

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

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

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

$\textsf{Making the denominators equal}$

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

$\mathsf{=\dfrac{-xy+xz-yz+xy-zx+yz}{(x-y)(y-z)(z-x)}}$

$\mathsf{=\dfrac{0}{(x-y)(y-z)(z-x)}}$

$\mathsf{=0}$

$\implies\boxed{\mathsf{\dfrac{x}{(x-y)(x-z)}+\dfrac{y}{(y-z)(y-x)}+\dfrac{z}{(z-x)(z-y)}=0}}$