If w is a complex cube root of unity, then the value of the determinant
`Delta = [(1,w,w^(2)),(w,w^(2),1),(w^(2),1,w)]`, is

Options

A. 1
B. 0
C. `w^(2)`
D. w

Explanation

To find the value of the determinant

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

where \( w \) is a complex cube root of unity, we can follow these steps:

### Step 1: Understand the properties of cube roots of unity
The complex cube roots of unity are \( 1, w, w^2 \) where \( w = e^{2\pi i / 3} \) and satisfies the equation \( w^3 = 1 \). This also implies that:

\[ 1 + w + w^2 = 0 \]

### Step 2: Apply a column operation
We can simplify the determinant by applying a column operation. We will add all three columns together to the first column:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ C_1 = 1 + w + w^2 = 0 \]

So, the first column becomes:

\[ C_1 = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

### Step 3: Write the new determinant
After performing the column operation, the determinant now looks like:

\[ \Delta = \begin{vmatrix} 0 & w & w^2 \\ 0 & w^2 & 1 \\ 0 & 1 & w \end{vmatrix} \]

### Step 4: Evaluate the determinant
Since the first column is entirely zeros, the value of the determinant is:

\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = 0 \]