If `alpha` and `beta` are the roots of the equation `ax^2 + bx +c =0 (a != 0; a, b,c` being different), then `(1+ alpha + alpha^2) (1+ beta+ beta^2) =`

Options

A. zero
B. positive
C. negative
D. none of these

Answer

To solve the problem, we need to evaluate the expression \((1 + \alpha + \alpha^2)(1 + \beta + \beta^2)\) given that \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(ax^2 + bx + c = 0\).

### Step-by-Step Solution:

1. **Identify the roots and their properties**:
From Vieta's formulas, for the quadratic equation \(ax^2 + bx + c = 0\):
- The sum of the roots \(\alpha + \beta = -\frac{b}{a}\) (Equation 1)
- The product of the roots \(\alpha \beta = \frac{c}{a}\) (Equation 2)

2. **Expand the expression**:
We need to expand \((1 + \alpha + \alpha^2)(1 + \beta + \beta^2)\):
\[ (1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = 1 \cdot 1 + 1 \cdot \beta + 1 \cdot \beta^2 + \alpha \cdot 1 + \alpha \cdot \beta + \alpha \cdot \beta^2 + \alpha^2 \cdot 1 + \alpha^2 \cdot \beta + \alpha^2 \cdot \beta^2 \]
This simplifies to:
\[ 1 + \beta + \beta^2 + \alpha + \alpha\beta + \alpha\beta^2 + \alpha^2 + \alpha^2\beta + \alpha^2\beta^2 \]

3. **Group the terms**:
Rearranging the terms gives:
\[ 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + (\alpha\beta + \alpha\beta^2 + \alpha^2\beta + \alpha^2\beta^2) \]

4. **Use identities**:
We know that:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting from Equations 1 and 2:
\[ \alpha^2 + \beta^2 = \left(-\frac{b}{a}\right)^2 - 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} - \frac{2c}{a} \]

5. **Substitute back into the expression**:
Now, substituting back into our expression:
\[ 1 + \left(-\frac{b}{a}\right) + \left(\frac{b^2}{a^2} - \frac{2c}{a}\right) + \left(\frac{c}{a} + \alpha\beta^2 + \alpha^2\beta + \alpha^2\beta^2\right) \]

6. **Final simplification**:
Collecting all terms gives:
\[ 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{2c}{a} + \frac{c}{a} \]
This results in:
\[ 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} \]

7. **Conclusion**:
The expression simplifies to a positive value because the squares of real numbers are non-negative, and since \(a\), \(b\), and \(c\) are different and \(a \neq 0\), the entire expression cannot equal zero or negative.

### Final Answer:
The expression \((1 + \alpha + \alpha^2)(1 + \beta + \beta^2)\) is always positive.