If $$x=\frac{a-b}{a+b},y=\frac{b-c}{b+c},z=\frac{c-a}{c+a}$$ then $$\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$$ is equal to

If $$x=\frac{a-b}{a+b}$$

=> $$(1-x) = 1- (\frac{a-b}{a+b})$$

=> $$(1-x) = \frac{2b}{a+b}$$

Similarly, $$(1+x) = \frac{2a}{a+b}$$

Applying the same method, we get :

=> $$(1-y) = \frac{2c}{b+c}$$ and => $$(1+y) = \frac{2b}{b+c}$$

=> $$(1-z) = \frac{2a}{c+a}$$ and => $$(1+z) = \frac{2c}{c+a}$$

Putting above values in the equation : $$\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}$$

=> $$\frac{(\frac{2b}{a+b})(\frac{2c}{b+c})(\frac{2a}{c+a})}{(\frac{2a}{a+b})(\frac{2b}{b+c})(\frac{2c}{c+a})}$$

=> $$\frac{2a*2b*2c}{2a*2b*2c}$$

= 1