If ` x^(2) = y + z, y^(2) = z + x, z^(2) = x + y` then the value of `(1)/( x + 1) + (1)/( y + 1) + (1)/( z+ 1)` is

Options

A. A)`-1`
B. B)1
C. C)2
D. D)4

Answer

To solve the problem, we start with the given equations:

1. \( x^2 = y + z \)
2. \( y^2 = z + x \)
3. \( z^2 = x + y \)

We need to find the value of \( \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} \).

### Step 1: Assume \( x = y = z \)

Since the equations are symmetric in \( x, y, z \), we can assume \( x = y = z = k \).

### Step 2: Substitute into the first equation

Substituting \( x = y = z = k \) into the first equation:

\[ k^2 = k + k \]

This simplifies to:

\[ k^2 = 2k \]

### Step 3: Rearrange the equation

Rearranging gives us:

\[ k^2 - 2k = 0 \]

### Step 4: Factor the equation

Factoring out \( k \):

\[ k(k - 2) = 0 \]

### Step 5: Solve for \( k \)

This gives us two solutions:

1. \( k = 0 \)
2. \( k = 2 \)

### Step 6: Choose a valid solution

Since \( k = 0 \) would lead to division by zero in our final expression, we choose \( k = 2 \). Thus, we have:

\[ x = 2, \quad y = 2, \quad z = 2 \]

### Step 7: Substitute back to find the desired expression

Now, substituting \( x, y, z \) into the expression we need to evaluate:

\[ \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} = \frac{1}{2 + 1} + \frac{1}{2 + 1} + \frac{1}{2 + 1} \]

This simplifies to:

\[ \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \]

### Final Answer

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