Two different real numbers `alpha and beta ` are the roots of the quadratic equation `ax ^(2) + c=0 a,c ne 0,` then ` alpha ^(3) + beta ^(3)` is:

Options

A. a
B. `-c`
C. 0
D. `-1`

Answer

To solve the problem, we need to find the value of \( \alpha^3 + \beta^3 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2 + c = 0 \) where \( a \) and \( c \) are not equal to zero.

### Step-by-step Solution:

1. **Identify the Roots**:
The given quadratic equation is \( ax^2 + c = 0 \). We can rewrite it as:
\[ ax^2 = -c \]
Since \( a \neq 0 \) and \( c \neq 0 \), this equation has two roots \( \alpha \) and \( \beta \).

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a quadratic equation of the form \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

In our case, since \( b = 0 \) (the coefficient of \( x \) is zero), we have:
\[ \alpha + \beta = 0 \]
and
\[ \alpha \beta = \frac{c}{a} \]

3. **Express One Root in Terms of the Other**:
From \( \alpha + \beta = 0 \), we can express \( \beta \) in terms of \( \alpha \):
\[ \beta = -\alpha \]

4. **Calculate \( \alpha^3 + \beta^3 \)**:
We can use the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
Substituting \( \alpha + \beta = 0 \):
\[ \alpha^3 + \beta^3 = 0 \cdot (\alpha^2 - \alpha\beta + \beta^2) = 0 \]

5. **Conclusion**:
Therefore, the value of \( \alpha^3 + \beta^3 \) is:
\[ \alpha^3 + \beta^3 = 0 \]

### Final Answer:
\[ \alpha^3 + \beta^3 = 0 \]

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