$ ext{135. If } a+b+c=0 ext{, then the value of } \left( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} ight) ext{ is :}$

Question

$	ext{135. If } a+b+c=0 	ext{, then the value of } \left( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} ight) 	ext{ is :}$

Accepted Answer

To solve the problem and justify why Option 4 is correct, we need to evaluate the expression $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$ given that $a + b + c = 0$.

### Step-by-step Explanation:

1. **Substitute $c = -(a + b)$:**
   Since $a + b + c = 0$, we can express $c$ as $c = -(a + b)$.

2. **Substitute in the Expression:**
   The expression becomes:
   $$
   \frac{a^2}{b(-(a+b))} + \frac{b^2}{a(-(a+b))} + \frac{(-(a+b))^2}{ab}
   $$

3. **Simplify Each Term:**
   - The first term: $\frac{a^2}{b(-(a+b))} = -\frac{a^2}{b(a+b)}$
   - The second term: $\frac{b^2}{a(-(a+b))} = -\frac{b^2}{a(a+b)}$
   - The third term: $\frac{(-(a+b))^2}{ab} = \frac{(a+b)^2}{ab}$

4. **Combine the Terms:**
   Combine the terms:
   $$
   -\frac{a^2}{b(a+b)} - \frac{b^2}{a(a+b)} + \frac{(a+b)^2}{ab}
   $$

5. **Common Denominator:**
   The common denominator for the first two terms is $ab(a+b)$. Rewrite the expression:
   $$
   = \frac{-a^3 - b^3 + (a+b)^3}{ab(a+b)}
   $$

6. **Expand $(a+b)^3$:**
   $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

7. **Substitute and Simplify:**
   Substitute and simplify:
   $$
   = \frac{-a^3 - b^3 + a^3 + 3a^2b + 3ab^2 + b^3}{ab(a+b)}
   $$
   $$
   = \frac{3a^2b + 3ab^2}{ab(a+b)}
   $$
   $$
   = \frac{3ab(a+b)}{ab(a+b)}
   $$
   $$
   = 3
   $$

### Conclusion:

The value of the expression is 3, which confirms that Option 4 is correct.

### Explanation of Incorrect Options:

- **Option 1 (0):** The expression simplifies to 3, not 0. The calculations clearly show that the terms do not cancel out to zero.
- **Option 2 (1):** The simplification process does not yield 1. The terms combine to give a result of 3.
- **Option 3 (-1):** The expression does not simplify to a negative value. The calculations show a positive result of 3.

Thus, the correct answer is indeed Option 4, as the expression evaluates to 3.