If `alpha, beta` are the roots of `px^(2) + qx + r = 0`, then `alpha^(3) + beta^(3) = "______"`.

Options

A. `(3qpr-q^(3))/(p^(3))`
B. `(3pqr-3q)/(p^(3))`
C. `(pqr - 3q)/(p^(3))`
D. `(3pqr -q)/(p^(3))`

Explanation

To find the value of \( \alpha^3 + \beta^3 \) given that \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( px^2 + qx + r = 0 \), we can use the following steps:

### Step 1: Identify the relationships between the roots and coefficients
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{q}{p} \)
- The product of the roots \( \alpha \beta = \frac{r}{p} \)

### Step 2: Use the identity for the sum of cubes
We can express \( \alpha^3 + \beta^3 \) using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We can also rewrite \( \alpha^2 + \beta^2 \) using the square of the sum of the roots:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting \( \alpha + \beta \) and \( \alpha \beta \):
\[ \alpha^2 + \beta^2 = \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) = \frac{q^2}{p^2} - \frac{2r}{p} \]

### Step 3: Substitute back into the sum of cubes formula
Now we can substitute \( \alpha^2 + \beta^2 \) into the identity for \( \alpha^3 + \beta^3 \):
\[ \alpha^3 + \beta^3 = (\alpha + \beta)\left(\alpha^2 + \beta^2 - \alpha\beta\right) \]
Substituting the values we have:
\[ \alpha^3 + \beta^3 = \left(-\frac{q}{p}\right)\left(\left(\frac{q^2}{p^2} - \frac{2r}{p}\right) - \frac{r}{p}\right) \]
Simplifying the expression inside the parentheses:
\[ \alpha^3 + \beta^3 = \left(-\frac{q}{p}\right)\left(\frac{q^2}{p^2} - \frac{3r}{p}\right) \]

### Step 4: Final simplification
Now we can simplify further:
\[ \alpha^3 + \beta^3 = -\frac{q}{p}\left(\frac{q^2 - 3rp}{p^2}\right) = -\frac{q(q^2 - 3rp)}{p^3} \]

Thus, the final answer is:
\[ \alpha^3 + \beta^3 = \frac{3rp - q^2}{p^2} \]