If `omega` is a cube root of unity, then Root of polynomial is
`|(x + 1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)|`

Options

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

Answer

To find the roots of the polynomial represented by the determinant

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

where \(\omega\) is a cube root of unity, we proceed as follows:

### Step 1: Set the determinant equal to zero
To find the roots of the polynomial, we set the determinant \(D\) equal to zero:
\[ D = 0 \]

### Step 2: Use properties of cube roots of unity
Recall that \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\). This will help simplify our calculations.

### Step 3: Apply row transformations
We can simplify the determinant by adding all rows together. Let’s perform the row operation \(R_1 \rightarrow R_1 + R_2 + R_3\):
\[ D = \begin{vmatrix} (x + 1) + (x + \omega^2) + (x + \omega) & \omega + 1 + 1 & \omega^2 + 1 + \omega \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{vmatrix} \]

### Step 4: Simplify using the properties of \(\omega\)
Using the property \(1 + \omega + \omega^2 = 0\), we can simplify the first row:
\[ D = \begin{vmatrix} 3x + 1 & 0 & 0 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{vmatrix} \]

### Step 5: Factor out common terms
Now, we can factor out \(x\) from the first row:
\[ D = x \begin{vmatrix} 1 & 0 & 0 \\ \omega & x + \omega^2 & 1 \\ \omega^2 & 1 & x + \omega \end{vmatrix} \]

### Step 6: Expand the determinant
Now, we can expand the determinant:
\[ D = x \left( 1 \cdot \begin{vmatrix} x + \omega^2 & 1 \\ 1 & x + \omega \end{vmatrix} \right) \]

Calculating the 2x2 determinant:
\[ \begin{vmatrix} x + \omega^2 & 1 \\ 1 & x + \omega \end{vmatrix} = (x + \omega^2)(x + \omega) - 1 \]

### Step 7: Set the determinant to zero
Setting the determinant equal to zero gives us:
\[ x \left( (x + \omega^2)(x + \omega) - 1 \right) = 0 \]

### Step 8: Solve for \(x\)
This gives us one root:
\[ x = 0 \]
Now, we need to solve the quadratic equation:
\[ (x + \omega^2)(x + \omega) - 1 = 0 \]

Expanding this:
\[ x^2 + (\omega + \omega^2)x + \omega^3 - 1 = 0 \]
Since \(\omega + \omega^2 = -1\) and \(\omega^3 = 1\):
\[ x^2 - x = 0 \]

### Step 9: Factor the quadratic
Factoring gives:
\[ x(x - 1) = 0 \]

### Step 10: Final roots
The roots are:
\[ x = 0, \quad x = 1 \]

### Conclusion
Thus, the roots of the polynomial are \(x = 0\) and \(x = 1\).

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