If `omega` is a complex cube root of unity, then a root of the equation
`|(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0`, is

Options

A. x = 1
B. `x = omega`
C. `x = omega^(2)`
D. `x = 0`

Explanation

To solve the given problem, we need to evaluate the determinant of the matrix and find the roots of the equation. The matrix is given as:

\[ \begin{vmatrix} x + 1 & \omega & \omega^2 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{vmatrix} = 0 \]

### Step 1: Write the determinant

We start with the determinant:

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

### Step 2: Simplify the determinant

To simplify the determinant, we can perform row operations. We can add the first row to the second and third rows:

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

This gives us:

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

### Step 3: Use properties of determinants

Next, we can factor out common terms or use properties of determinants. We can also use the fact that \(\omega + \omega^2 + 1 = 0\) (since \(\omega\) is a cube root of unity).

### Step 4: Expand the determinant

Now, we can expand the determinant using cofactor expansion. We will expand along the first row:

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

### Step 5: Calculate the 2x2 determinants

Calculating the 2x2 determinants:

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

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

3. \(\begin{vmatrix} \omega & x + \omega^2 \\ \omega^2 & 1 \end{vmatrix} = \omega - \omega^2(x + \omega^2) = \omega - \omega^2 x - \omega^4 = \omega - \omega^2 x - \omega\)

### Step 6: Substitute back into the determinant

Substituting these back into the determinant expression and simplifying gives us a polynomial in \(x\).

### Step 7: Solve for roots

Setting the determinant equal to zero gives us a polynomial equation in \(x\). We can solve this polynomial to find the roots.

### Final Result

After solving, we find that one of the roots is:

\[ x = 0 \]