Question

x+y+z=6 and xy+yz+zx=9 and xyz=1 then what is the value of (1/1-x)+(1/1-y)+(1/1-z)

Answer 1

Expanding, 
   { (1- x)(1-y) + (1-y)(1-z) + (1-z)(1-x) } / (1-x)(1-y)(1-z)
= { 1 - x - y + xy + 1 - y - z + yz - 1 - z - x + xz } / { ( 1 - x - y + xy )(1-z)
= { 3 - 2(x+y+z) + (xy + yz + zx) } / { 1 - x - y- z + xy + yz + xz - xyz }
= { 3 - 2*6 + 9 } / { 1 - 6 +9 -1 }
= (12 - 12) / 3 = 0 

There you go

Answer 2

Answer:

$<strong> </strong>\frac{1}{1 - x} + \frac{1}{1 - y} + \frac{1}{1 - z} =0$

Step-by-step explanation:

To solve

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

if

$x + y + z = 6 \\ \\ xy + yz + xz = 9 \\ \\ xyz = 1$

first take LCM

$\frac{(1 - y)(1 - z) + (1 -x )(1 - z) + (1 - x)(1 - y)}{(1 - x)(1 - y)(1 - z)} \\ \\ = > \frac{1 - z - y + yz + 1 - z - x + xz + 1 - y - x + xy}{(1 - x - y + xy)(1 - z)} \\ \\ = \frac{3 - 2x - 2y - 2z + xy + yz + xz}{1 - z - x + xz - y + yz + xy - xyz} \\ \\ \frac{3 - 2(x + y + z) + xy + yz + zx}{1 - (x + y + z) + xy + yz + xz - xyz } \\ \\ place \: all \: the \: values \: we \: have \\ \\ = \frac{3 - 2(6) + 9}{1 - (6) + 9 - 1} \\ \\ = \frac{3 - 12 + 9}{ - 6 + 9} \\ \\ = \frac{ 0}{3} \\ \\=0$

Hope it helps you.