Question

If xyz=1 ,then show that (1+x+y-1)-1 + (1+y+z-1)-1+(1+z+x-1)-1=1

Answer 1

hope this answer is helpful to you

Answer 2

The value of $\bold{\left(1+x+y^{-1}\right)^{-1}+\left(1+y+z^{-1}\right)^{-1}+\left(1+z+x^{-1}\right)^{-1}=1}$ is proven.

Solution:

Given:

xyz = 1

To prove: $\left(1+x+y^({-1}\right) )^{-1} +\left(1+y+z^({-1}\right) )^{-1} +\left(1+z+x^({-1}\right) )^{-1}$ =1

From xyz = 1, we can take $y=\frac{1}{z x}, x=\frac{1}{y z}, z=\frac{1}{x y}$  

Normally, the power -1 is considered as the denominator.

$\begin{array}{l}{=\frac{1}{1+x+\frac{1}{y}}+\frac{1}{1+y+\frac{1}{z}}+\frac{1}{1+z+\frac{1}{x}}} \\\\ {=\frac{y}{y+x y+1}+\frac{1}{1+y+x y}+\frac{1}{1+\frac{1}{x y}+\frac{1}{x}}} \\\\ {=\frac{y}{y+x y+1}+\frac{1}{1+y+x y}+\frac{x y}{1+y+x y}} \\ \\{=\frac{1+y+x y}{1+y+x y}=1}\end{array}$

Hence proved that the value of $\left(1+x+y^{-1}\right)^{-1}+\left(1+y+z^{-1}\right)^{-1}+\left(1+z+x^{-1}\right)^{-1}$  is 1 by considering xyz =1.