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 \(\frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)}\) given the definitions of \(x\), \(y\), and \(z\):

1. **Substituting the values of x, y, and z:**
- \(x = \frac{a-b}{a+b}\)
- \(y = \frac{b-c}{b+c}\)
- \(z = \frac{c-a}{c+a}\)

We will substitute these values into the expression.

2. **Calculating \(1-x\), \(1-y\), and \(1-z\):**
- \(1 - x = 1 - \frac{a-b}{a+b} = \frac{(a+b) - (a-b)}{a+b} = \frac{2b}{a+b}\)
- \(1 - y = 1 - \frac{b-c}{b+c} = \frac{(b+c) - (b-c)}{b+c} = \frac{2c}{b+c}\)
- \(1 - z = 1 - \frac{c-a}{c+a} = \frac{(c+a) - (c-a)}{c+a} = \frac{2a}{c+a}\)

3. **Calculating \(1+x\), \(1+y\), and \(1+z\):**
- \(1 + x = 1 + \frac{a-b}{a+b} = \frac{(a+b) + (a-b)}{a+b} = \frac{2a}{a+b}\)
- \(1 + y = 1 + \frac{b-c}{b+c} = \frac{(b+c) + (b-c)}{b+c} = \frac{2b}{b+c}\)
- \(1 + z = 1 + \frac{c-a}{c+a} = \frac{(c+a) + (c-a)}{c+a} = \frac{2c}{c+a}\)

4. **Substituting these into the expression:**
\[ \frac{(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)} = \frac{\left(\frac{2b}{a+b}\right) \left(\frac{2c}{b+c}\right) \left(\frac{2a}{c+a}\right)}{\left(\frac{2a}{a+b}\right) \left(\frac{2b}{b+c}\right) \left(\frac{2c}{c+a}\right)} \]

5. **Simplifying the expression:**
- The \(2\)s in the numerator and denominator cancel out.
- The expression simplifies to:
\[ \frac{bca}{abc} = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]