If `w ne 1` is a cube root of unity and `Delta=|{:(x+w^(2),w,1),(w,w^(2),1+x),(1,x+w,w^(2)):}|=0`, then value of `x` is

Options

A. `0`
B. `2`
C. `-1`
D. None of these

Explanation

To solve the given determinant problem, we start with the determinant defined as:

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

where \( \omega \) is a cube root of unity and \( \omega \neq 1 \).

### Step 1: Apply Column Transformation
We can simplify the determinant by applying the column transformation \( C_1 = C_1 + C_2 + C_3 \).

This gives us:

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

### Step 2: Simplify the First Column
Using the property of cube roots of unity, we know that \( 1 + \omega + \omega^2 = 0 \). Therefore, we can simplify the first column:

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

### Step 3: Expand the Determinant
Now we can expand the determinant. We will use the formula for a 3x3 determinant:

\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our determinant, we can denote:

- \( a = x + 1 \)
- \( b = \omega \)
- \( c = 1 \)
- \( d = x \)
- \( e = \omega^2 \)
- \( f = 1 + x \)
- \( g = x + 1 \)
- \( h = x + \omega \)
- \( i = \omega^2 \)

Calculating \( \Delta \):

\[ \Delta = (x + 1)(\omega^2(x + \omega) - (1 + x)(\omega^2)) - \omega(x(\omega^2) - (1 + x)(x + 1)) + 1(x(x + \omega) - \omega^2(x + 1)) \]

### Step 4: Set the Determinant to Zero
We are given that \( \Delta = 0 \). We need to find the values of \( x \) that satisfy this equation.

### Step 5: Test Possible Values
We can test the values \( x = 0, 2, -1 \) to see which one makes \( \Delta = 0 \).

1. **Testing \( x = 0 \)**:
\[ \Delta = \begin{vmatrix} 0 + \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 \\ 1 & 0 + \omega & \omega^2 \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 \\ 1 & \omega & \omega^2 \end{vmatrix} \]
This determinant evaluates to 0.

2. **Testing \( x = 2 \)** and **\( x = -1 \)**:
You can similarly substitute these values into the determinant and check if it equals 0.

After testing, we find that the only value that satisfies \( \Delta = 0 \) is:

\[ \boxed{0} \]