`|{:(1,omega,omega^2),(omega,omega^2,1),(omega^2,1,omega):}|`

Options

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

Answer

To solve the determinant

\[ D = \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \]

we will follow a systematic approach.

### Step 1: Rewrite the Determinant
We can denote the determinant as follows:

\[ D = \begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix} \]

### Step 2: Apply Column Operations
We will perform a column operation. We will replace the first column \(C_1\) with \(C_1 + C_2 + C_3\):

\[ C_1 \to C_1 + C_2 + C_3 \]

This gives us:

\[ D = \begin{vmatrix} 1 + \omega + \omega^2 & \omega & \omega^2 \\ \omega + \omega^2 + 1 & \omega^2 & 1 \\ \omega^2 + 1 + \omega & 1 & \omega \end{vmatrix} \]

### Step 3: Simplify the First Column
Now, we know that \(1 + \omega + \omega^2 = 0\) (this is a property of the cube roots of unity). Therefore, we can simplify the first column:

\[ D = \begin{vmatrix} 0 & \omega & \omega^2 \\ 0 & \omega^2 & 1 \\ 0 & 1 & \omega \end{vmatrix} \]

### Step 4: Evaluate the Determinant
Since the first column consists entirely of zeros, the determinant evaluates to zero:

\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is

\[ \boxed{0} \]

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