If `1,omega , omega^2` are the cube roots of unity , then `Delta=|(1,omega^n , omega^(2n)),(omega^n , omega^(2n), 1),(omega^(2n), 1, omega^n)|` is equal to :

Options

A. 0
B. 1
C. `omega`
D. `omega^2`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} \), where \( 1, \omega, \omega^2 \) are the cube roots of unity, we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} \]

### Step 2: Expand the Determinant
We can expand this determinant using the first row:
\[ \Delta = 1 \cdot \begin{vmatrix} \omega^{2n} & 1 \\ 1 & \omega^n \end{vmatrix} - \omega^n \cdot \begin{vmatrix} \omega^n & 1 \\ \omega^{2n} & \omega^n \end{vmatrix} + \omega^{2n} \cdot \begin{vmatrix} \omega^n & \omega^{2n} \\ \omega^{2n} & 1 \end{vmatrix} \]

### Step 3: Calculate Each 2x2 Determinant
1. The first determinant:
\[ \begin{vmatrix} \omega^{2n} & 1 \\ 1 & \omega^n \end{vmatrix} = \omega^{2n} \cdot \omega^n - 1 \cdot 1 = \omega^{3n} - 1 \]

2. The second determinant:
\[ \begin{vmatrix} \omega^n & 1 \\ \omega^{2n} & \omega^n \end{vmatrix} = \omega^n \cdot \omega^n - 1 \cdot \omega^{2n} = \omega^{2n} - \omega^{2n} = 0 \]

3. The third determinant:
\[ \begin{vmatrix} \omega^n & \omega^{2n} \\ \omega^{2n} & 1 \end{vmatrix} = \omega^n \cdot 1 - \omega^{2n} \cdot \omega^{2n} = \omega^n - \omega^{4n} \]

### Step 4: Substitute Back into the Determinant
Now substituting back into the expression for \( \Delta \):
\[ \Delta = 1 \cdot (\omega^{3n} - 1) - \omega^n \cdot 0 + \omega^{2n} \cdot (\omega^n - \omega^{4n}) \]
This simplifies to:
\[ \Delta = \omega^{3n} - 1 + \omega^{2n}(\omega^n - \omega^{4n}) = \omega^{3n} - 1 + \omega^{3n} - \omega^{6n} \]

### Step 5: Combine Like Terms
Combining the terms gives:
\[ \Delta = 2\omega^{3n} - 1 - \omega^{6n} \]

### Step 6: Use the Property of Cube Roots of Unity
Since \( \omega^3 = 1 \), we have \( \omega^{3n} = 1 \) and \( \omega^{6n} = 1 \):
\[ \Delta = 2 \cdot 1 - 1 - 1 = 2 - 1 - 1 = 0 \]

### Final Answer
Thus, the value of the determinant \( \Delta \) is:
\[ \Delta = 0 \]