If `alpha , beta ` are roots `x^(2) + px + q = 0` then value of `alpha^(3) + beta^(3)` is

Options

A. `3 pq+ p ^(3)`
B. `3pq - p^(3)`
C. 3pq
D. `p^(3) - 3 pq`

Answer

To find the value of \( \alpha^3 + \beta^3 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + px + q = 0 \), we can follow these steps:

### Step 1: Use the identity for the sum of cubes
We know that:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
This identity will help us express \( \alpha^3 + \beta^3 \) in terms of \( \alpha + \beta \) and \( \alpha \beta \).

### Step 2: Find \( \alpha + \beta \) and \( \alpha \beta \)
From Vieta's formulas, for the quadratic equation \( x^2 + px + q = 0 \):
- The sum of the roots \( \alpha + \beta = -p \)
- The product of the roots \( \alpha \beta = q \)

### Step 3: Calculate \( \alpha^2 + \beta^2 \)
We can express \( \alpha^2 + \beta^2 \) using the identity:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (-p)^2 - 2q = p^2 - 2q \]

### Step 4: Substitute into the sum of cubes formula
Now, we substitute \( \alpha + \beta \) and \( \alpha^2 + \beta^2 \) into the sum of cubes formula:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha\beta) \]
Substituting the values we have:
\[ \alpha^3 + \beta^3 = (-p)((p^2 - 2q) - q) \]
This simplifies to:
\[ \alpha^3 + \beta^3 = -p(p^2 - 3q) \]

### Final Result
Thus, the value of \( \alpha^3 + \beta^3 \) is:
\[ \alpha^3 + \beta^3 = -p(p^2 - 3q) \]