`a ^(2) = b + c, b ^(2) = c + a, c ^(2) = a + b,` then the value of `3 ((1)/( a + 1) + (1)/(b + 1) + (1)/( c +1))`

Options

A. 1
B. `1/3`
C. 3
D. 4

Answer

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

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

We need to find the value of \( 3 \left( \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \right) \).

### Step 1: Analyze the equations
From the symmetry of the equations, we can assume that \( a = b = c \). Let’s denote this common value as \( k \).

### Step 2: Substitute into the equations
If \( a = b = c = k \), then we can substitute \( k \) into the first equation:

\[ k^2 = k + k \]
This simplifies to:

\[ k^2 = 2k \]

### Step 3: Rearranging 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. \( k = 2 \)

Since \( a, b, c \) must be positive, we take \( k = 2 \). Therefore, we have:

\[ a = b = c = 2 \]

### Step 6: Substitute back into the expression
Now we substitute \( a, b, c \) into the expression we need to evaluate:

\[ 3 \left( \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} \right) = 3 \left( \frac{1}{2+1} + \frac{1}{2+1} + \frac{1}{2+1} \right) \]

### Step 7: Simplify the expression
Calculating \( \frac{1}{2+1} \):

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

Thus, we have:

\[ 3 \left( \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \right) = 3 \left( \frac{3}{3} \right) = 3 \]

### Final Answer
The value is \( 3 \).

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