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 problem, we start with the determinant given by:

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

### Step 1: Use the property of cube roots of unity

Recall that for cube roots of unity, \(1 + \omega + \omega^2 = 0\). This property will help simplify our calculations.

### Step 2: Simplify the determinant

We can simplify the determinant by performing row operations. We will add all three rows together:

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

This gives us:

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

Calculating the first row:

\[ x + \omega^2 + \omega + 1 = x + 0 = x \]

So we have:

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

### Step 3: Expand the determinant

Now we can expand the determinant using the first row:

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

### Step 4: Calculate the 2x2 determinants

1. For the first determinant:

\[ \begin{vmatrix} \omega^2 & 1 + x \\ x + \omega & \omega^2 \end{vmatrix} = \omega^2 \cdot \omega^2 - (1 + x)(x + \omega) = \omega^4 - (1 + x)(x + \omega) \]

2. For the second determinant:

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

3. For the third determinant:

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

### Step 5: Substitute back into the determinant

Now substituting back into the determinant:

\[ \Delta = x(\omega^4 - (1 + x)(x + \omega)) + \omega x - \omega^2 x \]

Since \(\omega^4 = \omega\), we simplify:

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

### Step 6: Set the determinant to zero

Now we set \(\Delta = 0\):

\[ x(\omega - (1 + x)(x + \omega)) + \omega x - \omega^2 x = 0 \]

### Step 7: Solve for \(x\)

We can factor out \(x\):

\[ x(\omega - (1 + x)(x + \omega) + \omega - \omega^2) = 0 \]

This gives us two cases:

1. \(x = 0\)
2. The expression in the parentheses equals zero.

Since we are looking for values of \(x\) that satisfy the determinant being zero, we find that:

\[ x = 0 \]

Thus, the value of \(x\) is:

\[ \boxed{0} \]