If `alpha, beta and gamma` are the roots of the equation `x^(3) + px + q = 0` (with `p != 0 and p != 0 and q != 0`), the value of the determinant
`|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|`, is

Options

A. p
B. q
C. `p^(2) - 2q`
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant:

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

### Step 1: Write down the determinant
We start with the determinant as given:

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

### Step 2: Use properties of determinants
We can perform row operations on the determinant. Specifically, we can add all three rows together and replace the first row with this sum:

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives us:

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

### Step 3: Simplify the determinant
Now, notice that the first row consists of the same element repeated three times. Therefore, we can factor out \((\alpha + \beta + \gamma)\):

\[ D = (\alpha + \beta + \gamma) \begin{vmatrix} 1 & 1 & 1 \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \]

### Step 4: Find the sum of the roots
From the given polynomial \(x^3 + px + q = 0\), we know that the sum of the roots \(\alpha + \beta + \gamma\) is given by:

\[ \alpha + \beta + \gamma = -\frac{b}{a} = 0 \]

since there is no \(x^2\) term in the polynomial.

### Step 5: Substitute the sum into the determinant
Substituting \(\alpha + \beta + \gamma = 0\) into our determinant:

\[ D = 0 \cdot \begin{vmatrix} 1 & 1 & 1 \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{0} \]