Question

If $x+y+z=0$,then prove that $\frac{xyz}{(x+y)(y+z)(z+x)}=-1$ where ($x\ne -y,y\ne -z,z\ne -x$) ?

Answer 1

$x+y+z=0$

then, $\frac{xyz}{(x+y)(y+z)(z+x)}=-1$

LHS. $\frac{xyz}{(x+y)(y+z)(z+x)}$

$\Rightarrow \frac{xyz}{xy+y^2+xz+yz\left)\right(z+x)}$

$\Rightarrow \frac{xyz}{xyz+y^{2}z+x{z}^2+z^{2}y+x^{2}y+x{y}^2+x^{2}z+xyz}$

$\Rightarrow \frac{xyz}{2xyz+y^{2}z+x{z}^2+z^{2}y+x^{2}y+x{y}^2+x^{2}z}$

$\Rightarrow \frac{xyz}{2xyz+y^{2}z+z^{2}y+x{z}^2+x^{2}z+x^{2}y+x{y}^2}$

$\Rightarrow \frac{xyz}{2xyz+yz(z+y)+xz(z+x)+xy(x+y)}-----\left(1\right)$

$\begin{array}{c}\therefore x+y+z=0\\ z+x=-y\\ x+y=-z\\ z+y=-x\end{array}]$

$\Rightarrow \frac{xyz}{2xyz+yz(-x)+xz(-y)+xy(-z)}$

$\Rightarrow \frac{xyz}{2xyz+(-xyz-xyz-xyz)}$

$\Rightarrow \frac{xyz}{2xyz-3xyz}=\frac{xyz}{-xyz}$

$=-1$ RHS.