Statement -1 : if a,b,c not all equal and ` a ne 0 ,a^(3) +b^(3) +c^(3) =3abc` ,then the equation `ax^(2) +bx+c=0` has two real roots of opposite sign.
and
Statement -2 : If roots of a quadractic equation ` ax^(2) +bx +c=0` are real and of opposite sign then `ac lt0.`

Options

A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -17
B. Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -17
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is Flase, Statement -2 is True

Answer

To solve the problem, we need to analyze both statements separately and determine their validity.

### Step 1: Analyze Statement 1
**Statement 1:** If \( a, b, c \) are not all equal, \( a \neq 0 \), and \( a^3 + b^3 + c^3 = 3abc \), then the equation \( ax^2 + bx + c = 0 \) has two real roots of opposite sign.

1. **Use the identity for cubes:**
The identity states:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Given \( a^3 + b^3 + c^3 = 3abc \), we can substitute this into the identity:
\[ 3abc - 3abc = 0 \]
This implies:
\[ (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]

2. **Analyze the two factors:**
Since \( a, b, c \) are not all equal, \( a + b + c = 0 \) must hold (the first factor must be zero).

3. **Substituting into the quadratic equation:**
The quadratic equation is \( ax^2 + bx + c = 0 \). If \( a + b + c = 0 \), we can substitute \( c = -a - b \).

4. **Finding the product of the roots:**
The product of the roots \( \alpha \) and \( \beta \) of the quadratic equation is given by:
\[ \alpha \beta = \frac{c}{a} = \frac{-a - b}{a} \]
The roots will be of opposite signs if \( \alpha \beta < 0 \).

5. **Condition for opposite signs:**
For \( \alpha \beta < 0 \), we need \( -a - b < 0 \) which implies \( a + b > 0 \). However, since \( a + b + c = 0 \), this means \( c < 0 \) when \( a + b > 0 \).

6. **Conclusion for Statement 1:**
We cannot guarantee that the roots are of opposite signs because the conditions can lead to both roots being positive or both being negative. Therefore, **Statement 1 is false**.

### Step 2: Analyze Statement 2
**Statement 2:** If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and of opposite sign, then \( ac < 0 \).

1. **Roots of opposite sign condition:**
If the roots \( \alpha \) and \( \beta \) are of opposite signs, then:
\[ \alpha \beta < 0 \]
The product of the roots is given by:
\[ \alpha \beta = \frac{c}{a} \]
Therefore, \( \frac{c}{a} < 0 \) implies:
\[ c \cdot a < 0 \]
This means that one of \( a \) or \( c \) must be positive and the other must be negative.

2. **Conclusion for Statement 2:**
Thus, if the roots are real and of opposite sign, it follows that \( ac < 0 \). Therefore, **Statement 2 is true**.

### Final Conclusion:
- **Statement 1 is false.**
- **Statement 2 is true.**