If `x=(a-b)/(a+b),y=(b-c)/(b+c),z=(c-a)/(c+a)` ,then the value of `((1+x)(1+y)(1+z))/((1-x)(1-y)(1-z))` is

Options

A. `abc`
B. `a^(2)b^(2)c^(2)`
C. `1`
D. `-1`

Answer

To find the value of \(\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}\) where \(x=\frac{a-b}{a+b}\), \(y=\frac{b-c}{b+c}\), and \(z=\frac{c-a}{c+a}\), we will follow these steps:

### Step 1: Substitute the values of \(x\), \(y\), and \(z\)

We start by substituting the values of \(x\), \(y\), and \(z\) into the expression.

\[ 1+x = 1 + \frac{a-b}{a+b} = \frac{(a+b) + (a-b)}{a+b} = \frac{2a}{a+b} \]

\[ 1+y = 1 + \frac{b-c}{b+c} = \frac{(b+c) + (b-c)}{b+c} = \frac{2b}{b+c} \]

\[ 1+z = 1 + \frac{c-a}{c+a} = \frac{(c+a) + (c-a)}{c+a} = \frac{2c}{c+a} \]

### Step 2: Calculate the numerator

Now we can calculate the product of the numerators:

\[ (1+x)(1+y)(1+z) = \left(\frac{2a}{a+b}\right) \left(\frac{2b}{b+c}\right) \left(\frac{2c}{c+a}\right) = \frac{8abc}{(a+b)(b+c)(c+a)} \]

### Step 3: Calculate \(1-x\), \(1-y\), and \(1-z\)

Next, we calculate \(1-x\), \(1-y\), and \(1-z\):

\[ 1-x = 1 - \frac{a-b}{a+b} = \frac{(a+b) - (a-b)}{a+b} = \frac{2b}{a+b} \]

\[ 1-y = 1 - \frac{b-c}{b+c} = \frac{(b+c) - (b-c)}{b+c} = \frac{2c}{b+c} \]

\[ 1-z = 1 - \frac{c-a}{c+a} = \frac{(c+a) - (c-a)}{c+a} = \frac{2a}{c+a} \]

### Step 4: Calculate the denominator

Now we can calculate the product of the denominators:

\[ (1-x)(1-y)(1-z) = \left(\frac{2b}{a+b}\right) \left(\frac{2c}{b+c}\right) \left(\frac{2a}{c+a}\right) = \frac{8abc}{(a+b)(b+c)(c+a)} \]

### Step 5: Form the final expression

Now we can substitute the values of the numerator and denominator into the expression:

\[ \frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)} = \frac{\frac{8abc}{(a+b)(b+c)(c+a)}}{\frac{8abc}{(a+b)(b+c)(c+a)}} = 1 \]

### Final Answer

Thus, the value of \(\frac{(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}\) is:

\[ \boxed{1} \]