If ab + bc + ca = 0, then find the value of `(b)/((a+b)(b+c))+(c)/((a+c)(c+b))+(a)/((a+b)(c+a))`

Options

A. 1
B. 0
C. `-1`
D. 2

Answer

To solve the problem, we need to find the value of the expression

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

given that \( ab + bc + ca = 0 \).

### Step 1: Write down the expression

We start with the expression:

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

### Step 2: Find a common denominator

The common denominator for the three fractions is:

\[ D = (a+b)(b+c)(c+a) \]

### Step 3: Rewrite each term with the common denominator

Now we can rewrite each term with the common denominator:

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

### Step 4: Combine the fractions

Combine the fractions over the common denominator:

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

### Step 5: Expand the numerator

Now we expand the numerator:

\[ b(c+a) + c(a+b) + a(b+c) = bc + ba + ca + cb + ab + ac \]

### Step 6: Group the terms

Notice that we can group the terms:

\[ = ab + ac + bc + ba + ca + cb = 2(ab + ac + bc) \]

### Step 7: Substitute the given condition

Since we know from the problem statement that \( ab + ac + bc = 0 \), we can substitute this into our expression:

\[ = 2(0) = 0 \]

### Step 8: Write the final answer

Thus, the value of the original expression \( S \) is:

\[ S = \frac{0}{(a+b)(b+c)(c+a)} = 0 \]

### Conclusion

The final answer is:

\[ \boxed{0} \]