Question

If xy + yz + zx = -1, then the value of (x+y/1+xy + z+y/1+zy + x+z/1+zx)​

Answer 1

$\frac{1}{xyz}$

Given that

xy + yz + zx = -1

Therefore

-1 = xy + yz + zx

1 = -(xy + yz + zx)

1 = -xy - yz - zx

To find, the value of

$\frac{x+y}{1+xy}$ + $\frac{z+y}{1+zy}$ + $\frac{x+z}{1+zx}$​

substitute -xy - yz - zx for 1

= $\frac{x+y}{-xy - yz - zx+xy}$ + $\frac{z+y}{-xy - yz - zx+zy}$ + $\frac{x+z}{-xy - yz - zx+zx}$

= $\frac{x+y}{-yz-zx}$ + $\frac{z+y}{-xy-zx}$ + $\frac{x+z}{-xy-yz}$

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

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

= -$\frac{1}{z}$-$\frac{1}{x}$-$\frac{1}{y}$

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

= -$\frac{xy + yz + zx}{xyz}$

= -$\frac{-1}{xyz}$

= $\frac{1}{xyz}$

The value of $\frac{x+y}{1+xy}+\frac{z+y}{1+zy}+\frac{x+z}{1+zx}$ is $\frac{1}{xyz}$