A) \[abc\]
B) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
C) 1
D) \[-1\]
Correct Answer: C
Solution :
We have, \[x=\frac{a-b}{a+b},y=\frac{b-c}{b+c},z=\frac{c-a}{c+a}\] Now, \[1+x=1+\frac{a-b}{a+b}=\frac{a+b+a-b}{a+b}=\frac{2a}{a+b}\] \[1-x=1-\frac{(a-b)}{a+b}=\frac{a+b-a+b}{a+b}=\frac{2b}{a+b}\] Similarly, \[1+y=\frac{2b}{b+c},1+z=\frac{2c}{a+c}\] \[1-y=\frac{2c}{b+c},1-z=\frac{2a}{a+c}\] Now, \[\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}\] \[=\frac{\left( \frac{2a}{a+b} \right)\left( \frac{2b}{b+c} \right)\left( \frac{2c}{a+c} \right)}{\left( \frac{2b}{a+b} \right)\left( \frac{2c}{b+c} \right)\left( \frac{2a}{a+c} \right)}=1\]
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