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 find the value of \( x \) such that the determinant

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

where \( \omega \) is a complex cube root of unity.

### Step 1: Use properties of cube roots of unity
Recall that for a complex cube root of unity, we have:

\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]

### Step 2: Simplify the determinant
We can simplify the determinant by adding all columns together. Let's add the first column, second column, and third column:

\[ C_1 \to C_1 + C_2 + C_3 \]

This gives us:

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

Using \( 1 + \omega + \omega^2 = 0 \), we find:

\[ x + 1 + 0 = x + 1 \]

Thus, the first column becomes:

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

### Step 3: Expand the determinant
Now we can expand the determinant along the first row:

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

### Step 4: Calculate the 2x2 determinants
Calculating the 2x2 determinants:

1. For the first determinant:

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

2. For the second determinant:

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

3. For the third determinant:

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

### Step 5: Set the determinant to zero
Setting \( D = 0 \) gives us a polynomial equation in \( x \). After simplifying and collecting like terms, we will find:

\[ x^3 - \omega = 0 \]

### Step 6: Solve for \( x \)
Thus, we find:

\[ x^3 = \omega \]

Taking the cube root gives us:

\[ x = 0 \]

### Final Answer
The root of the equation is:

\[ \boxed{0} \]