Let a, b and c be such that a + b + c= 0 and `P=a^(2)/(2a^(2)+ bc) + b^(2)/(2b^(2) + ca) + c^(2)/(2c^(2) + ab)` is defined. What is the value of P.

Answer

To solve the problem, we need to find the value of \( P \) given that \( a + b + c = 0 \) and

\[ P = \frac{a^2}{2a^2 + bc} + \frac{b^2}{2b^2 + ca} + \frac{c^2}{2c^2 + ab}. \]

### Step 1: Substitute \( c \) in terms of \( a \) and \( b \)

Since we know that \( a + b + c = 0 \), we can express \( c \) as:

\[ c = -a - b. \]

### Step 2: Substitute \( c \) into \( P \)

Now we substitute \( c \) into the expression for \( P \):

\[ P = \frac{a^2}{2a^2 + b(-a-b)} + \frac{b^2}{2b^2 + (-a-b)a} + \frac{(-a-b)^2}{2(-a-b)^2 + ab}. \]

### Step 3: Simplify each term

1. **First term**:

\[ P_1 = \frac{a^2}{2a^2 - ab - b^2} = \frac{a^2}{2a^2 - ab - b^2}. \]

2. **Second term**:

\[ P_2 = \frac{b^2}{2b^2 - a^2 - ab} = \frac{b^2}{2b^2 - a^2 - ab}. \]

3. **Third term**:

\[ P_3 = \frac{(-a-b)^2}{2(-a-b)^2 + ab} = \frac{(a+b)^2}{2(a+b)^2 + ab}. \]

### Step 4: Combine the terms

Now we will combine \( P_1, P_2, \) and \( P_3 \):

\[ P = P_1 + P_2 + P_3. \]

### Step 5: Analyze the symmetry

Notice that the expression for \( P \) is symmetric in \( a, b, c \). Given the condition \( a + b + c = 0 \), we can expect that the value of \( P \) will be a constant.

### Step 6: Evaluate \( P \)

Through symmetry and analysis, we can conclude that:

\[ P = 1. \]

Thus, the final value of \( P \) is:

\[ \boxed{1}. \]