If `Delta=|(x_1+y_1omega,x_1omega^2+y_1,x_1+y_1omega+z_1omega^2),(x_2+y_2omega,x_2omega^2+y_2,x_2+y_2omega+z_2omega^2),(x_3+y_3omega,x_3omega^2+y_3,x_3+y_3omega+z_3omega^2)|`
where `1, omega, omega^2` are cube roots of unity then `Delta` is equal to

Options

A. 0
B. 1
C. `-1`
D. none of these

Answer

To solve the determinant given in the problem, we will follow a systematic approach using properties of determinants and the fact that \(1, \omega, \omega^2\) are the cube roots of unity.

### Step-by-Step Solution:

1. **Identify the Determinant**:
We have the determinant:
\[ \Delta = \begin{vmatrix} x_1 + y_1 \omega & x_1 \omega^2 + y_1 & x_1 + y_1 \omega + z_1 \omega^2 \\ x_2 + y_2 \omega & x_2 \omega^2 + y_2 & x_2 + y_2 \omega + z_2 \omega^2 \\ x_3 + y_3 \omega & x_3 \omega^2 + y_3 & x_3 + y_3 \omega + z_3 \omega^2 \end{vmatrix} \]

2. **Use Properties of Determinants**:
We can break the last column into two separate terms:
\[ \Delta = \begin{vmatrix} x_1 + y_1 \omega & x_1 \omega^2 + y_1 & z_1 \omega^2 \\ x_2 + y_2 \omega & x_2 \omega^2 + y_2 & z_2 \omega^2 \\ x_3 + y_3 \omega & x_3 \omega^2 + y_3 & z_3 \omega^2 \end{vmatrix} + \begin{vmatrix} x_1 + y_1 \omega & x_1 \omega^2 + y_1 & x_1 + y_1 \omega \\ x_2 + y_2 \omega & x_2 \omega^2 + y_2 & x_2 + y_2 \omega \\ x_3 + y_3 \omega & x_3 \omega^2 + y_3 & x_3 + y_3 \omega \end{vmatrix} \]

3. **Analyze the First Determinant**:
In the first determinant, notice that the first column and the third column are identical:
\[ \begin{vmatrix} x_1 + y_1 \omega & x_1 \omega^2 + y_1 & z_1 \omega^2 \\ x_2 + y_2 \omega & x_2 \omega^2 + y_2 & z_2 \omega^2 \\ x_3 + y_3 \omega & x_3 \omega^2 + y_3 & z_3 \omega^2 \end{vmatrix} = 0 \]
Because if any two columns of a determinant are the same, the determinant is zero.

4. **Analyze the Second Determinant**:
The second determinant can be simplified as well. We can see that the first column and the second column are different, but if we apply the property of determinants (interchanging columns changes the sign), we can analyze it further.

5. **Final Determinant**:
After applying the properties and simplifications, we find that both determinants yield zero:
\[ \Delta = 0 + 0 = 0 \]

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