If ab + bc + ca = 0, then the value of `1/(a^2-bc) + 1/(b^2-ca) + 1/(c^2-ab)` will be:

Options

A. -1
B. a + b + c
C. abc
D. 0

Answer

To solve the problem \( ab + bc + ca = 0 \) and find the value of \( \frac{1}{a^2 - bc} + \frac{1}{b^2 - ca} + \frac{1}{c^2 - ab} \), we can follow these steps:

### Step 1: Analyze the given equation
We start with the equation:
\[ ab + bc + ca = 0 \]
This implies that \( ab = - (bc + ca) \).

### Step 2: Rewrite the denominators
We need to express \( a^2 - bc \), \( b^2 - ca \), and \( c^2 - ab \) in a useful form.

1. For \( a^2 - bc \):
\[ a^2 - bc = a^2 - (-ab - ca) = a^2 + ab + ca \]

2. For \( b^2 - ca \):
\[ b^2 - ca = b^2 - (-ab - bc) = b^2 + ab + bc \]

3. For \( c^2 - ab \):
\[ c^2 - ab = c^2 - (-bc - ca) = c^2 + bc + ca \]

### Step 3: Substitute back into the expression
Now we substitute these into the original expression:
\[ \frac{1}{a^2 + ab + ca} + \frac{1}{b^2 + ab + bc} + \frac{1}{c^2 + bc + ca} \]

### Step 4: Simplifying the expression
Since \( ab + bc + ca = 0 \), we can simplify each term:
1. For \( a^2 + ab + ca \):
\[ a^2 + ab + ca = a^2 - (bc) = a^2 - (-ab - ca) = a^2 + ab + ca = a^2 \]

2. For \( b^2 + ab + bc \):
\[ b^2 + ab + bc = b^2 - (ca) = b^2 - (-ab - bc) = b^2 + ab + bc = b^2 \]

3. For \( c^2 + bc + ca \):
\[ c^2 + bc + ca = c^2 - (ab) = c^2 - (-bc - ca) = c^2 + bc + ca = c^2 \]

### Step 5: Final expression
Now we have:
\[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \]

### Step 6: Conclusion
Thus, the value of \( \frac{1}{a^2 - bc} + \frac{1}{b^2 - ca} + \frac{1}{c^2 - ab} \) simplifies to:
\[ \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \]