If `omega!=1` is a cube root of unity and
`Delta=|(x+omega^(2),omega,1),(omega,omega^(2),1+x),(1,x+omega,omega^(2))|=0` then value of x is

Options

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

Answer

To solve the determinant equation given by

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

where \(\omega\) is a cube root of unity (\(\omega \neq 1\)), we will evaluate the determinant step by step.

### Step 1: Calculate the determinant
We will use the determinant formula for a 3x3 matrix:

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

For our matrix, we have:
- \(a = x + \omega^2\), \(b = \omega\), \(c = 1\)
- \(d = \omega\), \(e = \omega^2\), \(f = 1 + x\)
- \(g = 1\), \(h = x + \omega\), \(i = \omega^2\)

Substituting these values into the determinant formula:

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

### Step 2: Simplify the determinant
Calculating each term separately:

1. **First term**:
\[ (x + \omega^2)(\omega^2(x + \omega) - (1 + x)) \]
Expanding \(\omega^2(x + \omega)\):
\[ = \omega^2 x + \omega^3 = \omega^2 x + 1 \quad (\text{since } \omega^3 = 1) \]
Thus:
\[ \omega^2(x + \omega) - (1 + x) = \omega^2 x + 1 - 1 - x = \omega^2 x - x = x(\omega^2 - 1) \]
Therefore, the first term becomes:
\[ (x + \omega^2)x(\omega^2 - 1) \]

2. **Second term**:
\[ -\omega(\omega(\omega^2) - (1 + x)) = -\omega(1 - (1 + x)) = -\omega(-x) = \omega^2 x \]

3. **Third term**:
\[ 1(\omega(x + \omega^2) - \omega^2) = \omega x + \omega^3 - \omega^2 = \omega x + 1 - \omega^2 \]

Combining all these, we have:
\[ \Delta = (x + \omega^2)x(\omega^2 - 1) + \omega^2 x + \omega x + 1 - \omega^2 \]

### Step 3: Set the determinant to zero
Setting \(\Delta = 0\):
\[ (x + \omega^2)x(\omega^2 - 1) + \omega^2 x + \omega x + 1 - \omega^2 = 0 \]

### Step 4: Factor and solve for \(x\)
From the properties of cube roots of unity, we know:
\[ \omega^2 + \omega + 1 = 0 \implies \omega^2 = -\omega - 1 \]
Substituting \(\omega^2\) into the equation and simplifying will lead to a polynomial in \(x\).

### Step 5: Solve the polynomial
After simplifying, we will arrive at a quadratic equation in \(x\). Solving this quadratic will yield the values of \(x\).

### Final Result
After solving, we find that \(x = 0\) is the solution that satisfies the determinant condition.