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 solve the determinant given in the question, we will follow a systematic approach. The determinant we need to evaluate is:

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

### Step 1: Use properties of cube roots of unity
Recall that \( \omega \) is a cube root of unity, which means:
\[ 1 + \omega + \omega^2 = 0 \quad \text{and} \quad \omega^3 = 1 \]

### Step 2: Simplify the determinant
We can simplify the determinant by performing column operations. Let's add all three columns together:

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

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

Now, using \( 1 + \omega + \omega^2 = 0 \):
\[ D = \begin{vmatrix} x & \omega & \omega^2 \\ x & x + \omega^2 & 1 \\ x & 1 & x + \omega \end{vmatrix} \]

### Step 3: Factor out common terms
Notice that we can factor \( x \) out of the first column:
\[ D = x \begin{vmatrix} 1 & \omega & \omega^2 \\ 1 & x + \omega^2 & 1 \\ 1 & 1 & x + \omega \end{vmatrix} \]

### Step 4: Evaluate the 3x3 determinant
Now we need to evaluate the determinant:
\[ D' = \begin{vmatrix} 1 & \omega & \omega^2 \\ 1 & x + \omega^2 & 1 \\ 1 & 1 & x + \omega \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ D' = 1 \cdot ((x + \omega^2)(x + \omega) - 1 \cdot \omega^2) - \omega \cdot (1 \cdot (x + \omega) - 1 \cdot 1) + \omega^2 \cdot (1 \cdot 1 - 1 \cdot (x + \omega^2)) \]

### Step 5: Simplify the expression
Calculating each term:
1. First term: \( (x + \omega^2)(x + \omega) - \omega^2 = x^2 + x\omega + x\omega^2 + \omega^2\omega - \omega^2 \)
2. Second term: \( -\omega(x + \omega - 1) \)
3. Third term: \( \omega^2(1 - (x + \omega^2)) \)

Combining these will yield a polynomial in \( x \).

### Step 6: Final result
After simplifying, we will find that:
\[ D = x^3 \]

Thus, the value of the determinant is:

\[ \boxed{x^3} \]