If `alpha, beta, gamma` are the roots of `x^(3) + ax^(2) + b = 0`, then the value of `|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|`, is

Options

A. a
B. b
C. `a^(3)`
D. `b^(3)`

Answer

To solve the problem, we need to find the value of the determinant

\[ D = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \]

where \(\alpha, \beta, \gamma\) are the roots of the polynomial \(x^3 + ax^2 + b = 0\).

### Step-by-step Solution:

1. **Understanding the Determinant**:
The determinant \(D\) can be computed using the properties of determinants. This specific form is known as a cyclic determinant.

2. **Using the Cyclic Property**:
The value of the determinant can be expressed in terms of the roots. For a cyclic determinant of this form, there is a known result:
\[ D = 3\alpha\beta\gamma - (\alpha^3 + \beta^3 + \gamma^3) \]

3. **Finding \(\alpha^3 + \beta^3 + \gamma^3\)**:
We can use the identity:
\[ \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) \]
Here, we know:
- \(\alpha + \beta + \gamma = -a\) (from Vieta's formulas)
- \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\) (since \(b = 0\) in the polynomial)

4. **Calculating \(\alpha^2 + \beta^2 + \gamma^2\)**:
We can express \(\alpha^2 + \beta^2 + \gamma^2\) using:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the known values:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-a)^2 - 2(0) = a^2 \]

5. **Substituting Back**:
Now we substitute back into the identity for \(\alpha^3 + \beta^3 + \gamma^3\):
\[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha\beta\gamma + (-a)(a^2) \]
Since \(\alpha\beta\gamma = -b\), we can write:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3(-b) - a^3 \]

6. **Final Calculation**:
Substitute \(\alpha^3 + \beta^3 + \gamma^3\) back into the determinant:
\[ D = 3\alpha\beta\gamma - (3(-b) - a^3) = 3(-b) + a^3 \]
Since \(b = 0\) (as given in the polynomial), we find:
\[ D = a^3 \]

7. **Conclusion**:
Thus, the value of the determinant is:
\[ D = 3a^3 \]

### Final Answer:
The value of \(|(\alpha, \beta, \gamma), (\beta, \gamma, \alpha), (\gamma, \alpha, \beta)|\) is \(3a^3\).