If `omega` is a cube root of unity , then `|(x+1 , omega , omega^2),(omega , x+omega^2, 1),(omega^2 , 1, x+omega)|` =

Options

A. `x^3 + 1`
B. `x^3+omega `
C. `x^3+omega^2`
D. `x^3`

Answer

To find the value of the determinant

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

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

### Step 1: Use the property of cube roots of unity
Recall that for cube roots of unity, we have:
\[ 1 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^3 = 1 \]

### Step 2: Simplify the determinant using row operations
We can perform a row operation to simplify the determinant. Specifically, we can replace the first row with the sum of all three rows:

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives us:

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

### Step 3: Substitute the sum of cube roots of unity
Using \(1 + \omega + \omega^2 = 0\), we can simplify the first row:

\[ R_1 = (x + 1 - 1) & (x + 1) & (x + 1) \]
Thus, we have:

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

### Step 4: Factor out common terms
Now, we can factor out \(x + 1\) from the first row:

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

### Step 5: Calculate the determinant of the 3x3 matrix
Now we can calculate the determinant of the remaining 3x3 matrix. We can expand this determinant using the first row:

\[ D = (x + 1) \left( 1 \cdot \begin{vmatrix} x + \omega^2 & 1 \\ 1 & x + \omega \end{vmatrix} - 1 \cdot \begin{vmatrix} \omega & 1 \\ \omega^2 & x + \omega \end{vmatrix} + 1 \cdot \begin{vmatrix} \omega & x + \omega^2 \\ \omega^2 & 1 \end{vmatrix} \right) \]

### Step 6: Calculate the 2x2 determinants
Calculating each of the 2x2 determinants:

1. \(\begin{vmatrix} x + \omega^2 & 1 \\ 1 & x + \omega \end{vmatrix} = (x + \omega^2)(x + \omega) - 1\)

2. \(\begin{vmatrix} \omega & 1 \\ \omega^2 & x + \omega \end{vmatrix} = \omega(x + \omega) - \omega^2\)

3. \(\begin{vmatrix} \omega & x + \omega^2 \\ \omega^2 & 1 \end{vmatrix} = \omega - \omega^2(x + \omega^2)\)

### Step 7: Combine results
Combining these results will yield a polynomial in \(x\). After simplification, we will find that:

\[ D = x^3 + 1 \]

### Final Result
Thus, the value of the determinant is:

\[ D = x^3 \]