Question

Given $a=\frac{x}{y-z};b=\frac{y}{z-x};c=\frac{z}{x-y}$, where $x,y,z$ are not all zero, prove that $:1+ab+bc+ca$.

Answer 1

Given:-

$a=\frac{x}{y-z}$

$b=\frac{y}{z-x}$

$c=\frac{z}{x-y}$

To prove:- $1+ab+bc+ca=0$

Proof:-

$ab=\frac{x}{y-z}\times \frac{y}{z-x}=\frac{xy}{(y-z)(z-x)}$

$bc=\frac{y}{z-x}\times zx-y=\frac{yz}{(z-x)(x-y)}$

$ca=\frac{z}{x-y}\times \frac{x}{y-z}=\frac{xz}{(x-y)(y-z)}$

$1+ab+bc+ca=0$

$1+\frac{xy}{(y-z)(z-x)}+\frac{yz}{(z-x)(x-y)}+\frac{xz}{(x-y)(y-z)}=0$

$1+\frac{xy(x-y)+yz(y-z)+zx(z-x)}{(x-y)(y-z)(z-x)}=0$

$1+\frac{x^{2}y-x{y}^2+y^{2}z-y{z}^2+zx(z-x)}{(x-y)(y-z)(z-x)}=0$

$1+\frac{y(x^2-z^2)+y^2(z-x)+zx(z-x)}{(x-y)(y-z)(z-x)}=0$

$1+\frac{(z-x)(-y(x+z)+y^2+zx)}{(x-y)(y-z)(z-x)}=0$

$1+\frac{(z-x)(y^2-xy-yz+zx)}{(x-y)(y-z)(z-x)}=0$

$1+\frac{(z-x)(y-x)(y-z)}{(x-y)(y-z)(z-x)}=0$

$1+(-1)=0$

$1-1=0$

L.H.S. $=$ R.H.S.

Hence proved that $1+ab+bc+ca=0$.