If ` x = (a - b)/( a + b) , y = (b - c)/( b + c) , z = ( c - a)/( c +a)` ,then `((1 - x) (1 - y) (1 - z))/( (1 + x) (1 + y) (1 + z))`is equal to

Options

A. 1
B. 0
C. 2
D. `(1)/(2)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{(1 - x)(1 - y)(1 - z)}{(1 + x)(1 + y)(1 + z)} \]

where

\[ x = \frac{a - b}{a + b}, \quad y = \frac{b - c}{b + c}, \quad z = \frac{c - a}{c + a} \]

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

First, we calculate \(1 - x\):

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

Next, we calculate \(1 - y\):

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

Finally, we calculate \(1 - z\):

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

### Step 2: Calculate \(1 + x\), \(1 + y\), and \(1 + z\)

Now we calculate \(1 + x\):

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

Next, we calculate \(1 + y\):

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

Finally, we calculate \(1 + z\):

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

### Step 3: Substitute into the expression

Now we substitute these results into the original expression:

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

### Step 4: Simplify the expression

The expression simplifies as follows:

\[ = \frac{2b \cdot 2c \cdot 2a}{2a \cdot 2b \cdot 2c} \cdot \frac{(a + b)(b + c)(c + a)}{(a + b)(b + c)(c + a)} = 1 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{1} \]