if ` a^(2) = b + c, b^(2) = a + c, c^(2) = b + a ` then what will be the value of ` (1)/ ( a + 1) + (1)/( b + 1) + (1)/( c + 1)`

Options

A. a)`-1`
B. b)2
C. c)1
D. d)0

Answer

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

1. \( a^2 = b + c \)
2. \( b^2 = a + c \)
3. \( c^2 = a + b \)

We need to find the value of:

\[ \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \]

### Step 1: Assume values for \( a, b, c \)

To simplify the problem, we can assume values for \( a, b, \) and \( c \). Let's assume:

\[ a = 2, \quad b = 2, \quad c = 2 \]

### Step 2: Verify the assumptions

Now we check if these values satisfy the original equations:

- For \( a^2 = b + c \):
\[ 2^2 = 2 + 2 \implies 4 = 4 \quad \text{(True)} \]

- For \( b^2 = a + c \):
\[ 2^2 = 2 + 2 \implies 4 = 4 \quad \text{(True)} \]

- For \( c^2 = a + b \):
\[ 2^2 = 2 + 2 \implies 4 = 4 \quad \text{(True)} \]

Since all equations hold true, our assumptions are valid.

### Step 3: Calculate the desired expression

Now we substitute \( a, b, c \) into the expression:

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

This simplifies to:

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

### Final Answer

Thus, the value of \( \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} \) is:

\[ \boxed{1} \]