If `alpha, beta, gamma` are the roots of `x^(3)+ax+b=0`, then find the value of `alpha^(3)+beta^(3)+gamma^(3)`.

Answer

To find the value of \( \alpha^3 + \beta^3 + \gamma^3 \) where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + ax + b = 0 \), we can use the relationship between the roots and coefficients of the polynomial.

### Step-by-step Solution:

1. **Identify the Coefficients:**
The given polynomial is \( x^3 + ax + b = 0 \). Here, we can identify:
- \( a_0 = 1 \) (coefficient of \( x^3 \))
- \( a_1 = 0 \) (coefficient of \( x^2 \))
- \( a_2 = a \) (coefficient of \( x \))
- \( a_3 = b \) (constant term)

2. **Use Vieta's Formulas:**
According to Vieta's formulas, we have:
- \( \alpha + \beta + \gamma = -\frac{a_1}{a_0} = -\frac{0}{1} = 0 \)
- \( \alpha \beta + \beta \gamma + \gamma \alpha = \frac{a_2}{a_0} = \frac{a}{1} = a \)
- \( \alpha \beta \gamma = -\frac{a_3}{a_0} = -\frac{b}{1} = -b \)

3. **Use the Identity for Sum of Cubes:**
We can use the identity:
\[ \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) \]
Since \( \alpha + \beta + \gamma = 0 \), the equation simplifies to:
\[ \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = 0 \]

4. **Rearranging the Equation:**
Rearranging gives us:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma \]

5. **Substituting the Value of \( \alpha\beta\gamma \):**
From Vieta's, we know \( \alpha\beta\gamma = -b \). Thus:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3(-b) = -3b \]

### Final Answer:
\[ \alpha^3 + \beta^3 + \gamma^3 = -3b \]