If `alpha,beta,gamma` are roots of the equation `x^(3)+px+q=0` then the value of `|{:(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta):}|` is

Options

A. p
B. q
C. `p^(2)-4q`
D. 0

Answer

To find the value of the determinant \( |(\alpha, \beta, \gamma), (\beta, \gamma, \alpha), (\gamma, \alpha, \beta)| \), where \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + px + q = 0 \), we can follow these steps:

### Step 1: Write down the determinant
The determinant can be expressed as:
\[ D = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \]

### Step 2: Use properties of determinants
We can use the property of determinants that states if two rows (or columns) of a determinant are identical, the determinant equals zero. In our case, we can manipulate the rows to find a simpler form.

### Step 3: Expand the determinant
We can expand the determinant using the first row:
\[ D = \alpha \begin{vmatrix} \gamma & \alpha \\ \alpha & \beta \end{vmatrix} - \beta \begin{vmatrix} \beta & \alpha \\ \gamma & \beta \end{vmatrix} + \gamma \begin{vmatrix} \beta & \gamma \\ \gamma & \alpha \end{vmatrix} \]

### Step 4: Calculate the 2x2 determinants
Calculating the 2x2 determinants:
1. \( \begin{vmatrix} \gamma & \alpha \\ \alpha & \beta \end{vmatrix} = \gamma \beta - \alpha^2 \)
2. \( \begin{vmatrix} \beta & \alpha \\ \gamma & \beta \end{vmatrix} = \beta^2 - \alpha \gamma \)
3. \( \begin{vmatrix} \beta & \gamma \\ \gamma & \alpha \end{vmatrix} = \beta \alpha - \gamma^2 \)

### Step 5: Substitute back into the determinant
Substituting these back into the expression for \( D \):
\[ D = \alpha (\gamma \beta - \alpha^2) - \beta (\beta^2 - \alpha \gamma) + \gamma (\beta \alpha - \gamma^2) \]

### Step 6: Simplify the expression
Now, simplifying the expression:
\[ D = \alpha \gamma \beta - \alpha^3 - \beta^3 + \alpha \beta \gamma + \beta \alpha \gamma - \gamma^3 \]
Combining the like terms, we get:
\[ D = 3\alpha \beta \gamma - (\alpha^3 + \beta^3 + \gamma^3) \]

### Step 7: Use the identity for the sum of cubes
Using the identity \( \alpha^3 + \beta^3 + \gamma^3 - 3\alpha \beta \gamma = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha) \) and since \( \alpha + \beta + \gamma = 0 \), we have:
\[ \alpha^3 + \beta^3 + \gamma^3 = 3\alpha \beta \gamma \]

### Step 8: Conclude the value of the determinant
Thus, substituting back, we find:
\[ D = 3\alpha \beta \gamma - 3\alpha \beta \gamma = 0 \]

### Final Answer
The value of the determinant \( |(\alpha, \beta, \gamma), (\beta, \gamma, \alpha), (\gamma, \alpha, \beta)| \) is:
\[ \boxed{0} \]