Question

If xyz = 1, then find the value of
(1+x+y^-1)^-1 + (1+y+z^-1)^-1 + (1+z+x^-1)^-1

Answer 1

$xyz=1\\xy=\frac{1}{z}=z^{-1},\ \ yz=\frac{1}{x}=x^{-1},\ \ \ zx=\frac{1}{y}=y^{-1}$

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

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

= 1
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you could try some quick calculations like:

let  x = 1 ,  y = 1  and z = 1         =>    A = 1/3 + 1/3 + 1/3 = 1
let  x = 1  , y = -1  and  z = -1      =>    A = 1/1 + 1/-1  + 1/1  = 1
let x = 2 , y = 1/2  and z = 1        =>    A = 1/5  + 1/2.5  + 1/2.5  = 1

So we can safely assume that the given complicated polynomial sum = 1

Answer 2

The value of $({1 + x + y ^{-1}})^{-1} + ({ 1+ y + z^{-1}}) ^{-1} + ({ 1 + z + x^{-1}})^{-1} = 1$

Given:

$x \times y \times z = 1$

To find:

The value of $({1 + x + y ^{-1}})^{-1} + ({ 1+ y + z^{-1}}) ^{-1} + ({ 1 + z + x^{-1}})^{-1} = 1$

Solution:

As $x \times y \times z=1$

x = 1, y = 1, z = 1

The power -1 is considered at the denominator here.

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

$= \frac {1}{3} + \frac {1}{3} + \frac {1}{3}$

$= \frac {3}{3}$

= 1

The value is 1.