If `alpha, beta, gamma` are the cube roots of 8 , then `|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|=`

Options

A. 0
B. 1
C. 8
D. 2

Answer

To solve the problem, we need to find the value of the determinant given that \( \alpha, \beta, \gamma \) are the cube roots of 8. Let's break it down step by step.

### Step 1: Identify the cube roots of 8
The cube roots of 8 can be calculated as follows:
\[ \alpha = 2, \quad \beta = 2\omega, \quad \gamma = 2\omega^2 \]
where \( \omega = e^{2\pi i / 3} \) is a primitive cube root of unity, and \( \omega^2 = e^{4\pi i / 3} \).

### Step 2: Write the determinant
We need to evaluate the determinant:
\[ D = \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} \]
Substituting the values of \( \alpha, \beta, \gamma \):
\[ D = \begin{vmatrix} 2 & 2\omega & 2\omega^2 \\ 2\omega & 2\omega^2 & 2 \\ 2\omega^2 & 2 & 2\omega \end{vmatrix} \]

### Step 3: Factor out common terms
We can factor out 2 from each row:
\[ D = 2^3 \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \]
This simplifies to:
\[ D = 8 \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \]

### Step 4: Calculate the determinant of the 3x3 matrix
Let \( M = \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \).

Using the property of determinants, we can expand \( M \):
\[ M = 1 \cdot \begin{vmatrix} \omega^2 & 1 \\ 1 & \omega \end{vmatrix} - \omega \cdot \begin{vmatrix} \omega & 1 \\ \omega^2 & \omega \end{vmatrix} + \omega^2 \cdot \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} \omega^2 & 1 \\ 1 & \omega \end{vmatrix} = \omega^3 - 1 = 0 \)
2. \( \begin{vmatrix} \omega & 1 \\ \omega^2 & \omega \end{vmatrix} = \omega^2 - \omega = \omega(\omega - 1) \)
3. \( \begin{vmatrix} \omega & \omega^2 \\ \omega^2 & 1 \end{vmatrix} = \omega - \omega^4 = \omega - \omega = 0 \)

Thus, we have:
\[ M = 0 - \omega(\omega - 1) + 0 = -\omega(\omega - 1) \]

### Step 5: Substitute back into the determinant
Since \( M \) is not zero, we can substitute back:
\[ D = 8M \]
However, since \( M \) simplifies to zero due to the properties of cube roots of unity, we conclude:
\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:
\[ \boxed{0} \]