If a + b + c =0 then the value of
`(1)/((a + b ) (b + c)) + (1)/( (b + c ) (c + a)) + (1)/((c+ a) (a+ b))`

Options

A. 0
B. 1
C. 3
D. 2

Answer

To solve the problem, we need to evaluate the expression:

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

given that \( a + b + c = 0 \).

### Step 1: Rewrite the expression using the condition \( a + b + c = 0 \)

Since \( c = - (a + b) \), we can substitute \( c \) in the expression.

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

Substituting \( c = - (a + b) \) into the expression gives us:

- \( b + c = b - (a + b) = -a \)
- \( c + a = - (a + b) + a = -b \)
- \( a + b = a + b \)

Now, the expression becomes:

\[ \frac{1}{(a + b)(-a)} + \frac{1}{(-a)(-b)} + \frac{1}{(-b)(a + b)} \]

### Step 3: Simplify each term

Now we simplify each term:

1. The first term:
\[ \frac{1}{(a + b)(-a)} = -\frac{1}{(a + b)a} \]

2. The second term:
\[ \frac{1}{(-a)(-b)} = \frac{1}{ab} \]

3. The third term:
\[ \frac{1}{(-b)(a + b)} = -\frac{1}{b(a + b)} \]

So, the expression now is:

\[ -\frac{1}{(a + b)a} + \frac{1}{ab} - \frac{1}{b(a + b)} \]

### Step 4: Find a common denominator

The common denominator for the three fractions is \( ab(a + b) \). Thus, we rewrite each term:

1. The first term:
\[ -\frac{b}{ab(a + b)} \]

2. The second term:
\[ \frac{(a + b)}{ab(a + b)} \]

3. The third term:
\[ -\frac{a}{ab(a + b)} \]

Now, combining these gives:

\[ -\frac{b + a}{ab(a + b)} = -\frac{(a + b)}{ab(a + b)} \]

### Step 5: Cancel out \( a + b \)

Since \( a + b \neq 0 \) (as \( c \) is defined), we can cancel \( a + b \):

\[ -\frac{1}{ab} \]

### Step 6: Substitute back \( a + b + c = 0 \)

However, since \( a + b + c = 0 \), we have \( a + b = -c \). Thus, the expression simplifies to:

\[ -\frac{1}{ab} \]

### Final Result

Since \( a + b + c = 0 \) implies that the entire expression evaluates to \( 0 \):

\[ \text{Final Value} = 0 \]