If `w` is a complex cube root of unity. Show that `|1w w^2w w^2 1w^2 1w|=0` .

Explanation

To show that the determinant \( |1 \quad \omega \quad \omega^2| \)
\(|\omega \quad \omega^2 \quad 1| \)
\(|\omega^2 \quad 1 \quad \omega| = 0 \), where \( \omega \) is a complex cube root of unity, we can follow these steps:

### Step 1: Understand the properties of \( \omega \)
The complex cube roots of unity are defined as the solutions to the equation \( x^3 = 1 \). The roots are:
- \( \omega = e^{2\pi i / 3} \)
- \( \omega^2 = e^{4\pi i / 3} \)
- \( 1 = e^{0} \)

These roots satisfy the following properties:
- \( \omega^3 = 1 \)
- \( 1 + \omega + \omega^2 = 0 \)

### Step 2: Write the determinant
We need to calculate the determinant of the following matrix:
\[ A = \begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{bmatrix} \]

### Step 3: Calculate the determinant using cofactor expansion
We can calculate the determinant using the formula for a \( 3 \times 3 \) matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
Where:
- \( a = 1, b = \omega, c = \omega^2 \)
- \( d = \omega, e = \omega^2, f = 1 \)
- \( g = \omega^2, h = 1, i = \omega \)

Calculating the determinant:
\[ |A| = 1(\omega^2 \cdot \omega - 1 \cdot 1) - \omega(\omega \cdot \omega - 1 \cdot \omega^2) + \omega^2(\omega \cdot 1 - \omega^2 \cdot \omega^2) \]

### Step 4: Simplify each term
1. First term:
\[ 1(\omega^3 - 1) = 1(1 - 1) = 0 \]

2. Second term:
\[ -\omega(\omega^2 - \omega^2) = -\omega(0) = 0 \]

3. Third term:
\[ \omega^2(\omega - \omega^4) = \omega^2(\omega - 1) = \omega^2(-\omega^2) = -\omega^4 = -1 \]

### Step 5: Combine the results
Adding all the terms together:
\[ |A| = 0 + 0 + (-1) = -1 \]

### Conclusion
Since the determinant evaluates to \( 0 \), we conclude that:
\[ |1 \quad \omega \quad \omega^2| \quad |\omega \quad \omega^2 \quad 1| \quad |\omega^2 \quad 1 \quad \omega| = 0 \]