Let `omega!=1` be a cube root of unit and
`Delta=|(1-omega-omega^(2),2,2),(2omega, omega-omega^(2)-1,2omega),(2omega^(2),2omega^(2),omega^(2)-1-omega)|`
then `Delta` equals

Options

A. `-omega`
B. `3omega(1-omega)`
C. `0`
D. `1-omega^(2)`

Answer

To find the value of \( \Delta \) given by the determinant

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

where \( \omega \) is a cube root of unity (i.e., \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \)), we can simplify the determinant using row operations.

### Step 1: Simplify the first row
We can replace the first row \( R_1 \) with \( R_1 + R_2 + R_3 \):

\[ R_1 = (1 - \omega - \omega^2) + (2\omega) + (2\omega^2) \]

Calculating this:

- The first element:
\[ 1 - \omega - \omega^2 + 2\omega + 2\omega^2 = 1 + \omega + \omega^2 = 1 + 0 = 1 \]

- The second element:
\[ 2 + (\omega - \omega^2 - 1) + 2\omega^2 = 2 + \omega - \omega^2 - 1 + 2\omega^2 = 1 + \omega + \omega^2 = 1 + 0 = 1 \]

- The third element:
\[ 2 + (\omega^2 - 1 - \omega) = 2 + \omega^2 - 1 - \omega = 1 + \omega^2 - \omega \]

Thus, the new first row becomes:

\[ R_1 = (1, 1, 1 + \omega^2 - \omega) \]

### Step 2: Rewrite the determinant
Now, the determinant becomes:

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

### Step 3: Identify a row of zeros
Next, we can see that \( 1 + \omega + \omega^2 = 0 \) implies \( \omega^2 = -1 - \omega \). Therefore, we can simplify the third row:

\[ \omega^2 - 1 - \omega = -1 - \omega - 1 - \omega = -2 - 2\omega \]

Thus, the third row can be simplified to:

\[ R_3 = (2\omega^2, 2\omega^2, -2 - 2\omega) \]

### Step 4: Calculate the determinant
Now, notice that if we replace \( R_3 \) with \( R_3 + 2R_1 \), we will have a row of zeros:

\[ R_3 = (2\omega^2 + 2, 2\omega^2 + 2, -2 - 2\omega + 2(1 + \omega^2 - \omega)) \]

This will lead to a row of zeros, making the determinant equal to zero:

\[ \Delta = 0 \]

### Final Answer
Thus, the value of \( \Delta \) is:

\[ \Delta = 0 \]