If ` alpha , beta , gamma ` are the roots of ` x^3 +px^2 +qx +r=0` then find
`sum alpha^3`

Answer

To find the sum of the cubes of the roots \( \alpha, \beta, \gamma \) of the polynomial equation \( x^3 + px^2 + qx + r = 0 \), we can use the relationships derived from Vieta's formulas and the identity for the sum of cubes.

### Step-by-Step Solution:

1. **Identify the Roots and Coefficients**:
Given the polynomial \( x^3 + px^2 + qx + r = 0 \), we identify:
- The roots are \( \alpha, \beta, \gamma \).
- Coefficients are \( a = 1, b = p, c = q, d = r \).

2. **Apply Vieta's Formulas**:
From Vieta's formulas, we have:
- \( \alpha + \beta + \gamma = -p \)
- \( \alpha \beta + \beta \gamma + \gamma \alpha = q \)
- \( \alpha \beta \gamma = -r \)

3. **Use the Identity for the Sum of Cubes**:
The sum of the cubes of the roots can be expressed as:
\[ \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha) + 3\alpha \beta \gamma \]

4. **Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)**:
We can find \( \alpha^2 + \beta^2 + \gamma^2 \) using the formula:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
Substituting the values:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-p)^2 - 2q = p^2 - 2q \]

5. **Substitute Back into the Sum of Cubes Formula**:
Now substitute \( \alpha + \beta + \gamma \) and \( \alpha^2 + \beta^2 + \gamma^2 \) into the sum of cubes formula:
\[ \alpha^3 + \beta^3 + \gamma^3 = (-p)((p^2 - 2q) - q) + 3(-r) \]
Simplifying this:
\[ = -p(p^2 - 3q) - 3r \]
\[ = -p^3 + 3pq - 3r \]

6. **Final Result**:
Therefore, the sum of the cubes of the roots is:
\[ \alpha^3 + \beta^3 + \gamma^3 = -p^3 + 3pq - 3r \]

### Summary:
The required value of \( \alpha^3 + \beta^3 + \gamma^3 \) is:
\[ \alpha^3 + \beta^3 + \gamma^3 = -p^3 + 3pq - 3r \]