Suppose `omega!=1` is a sube root of unity, and
`Delta=|(1,omega^(2),1-omega^(4)),(omega,1,1+omega^(5)),(1,omega,omega^(2))|` then `|Re(Delta)|=`_____________

Answer

To solve the problem, we need to compute the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} 1 & \omega^2 & 1 - \omega^4 \\ \omega & 1 & 1 + \omega^5 \\ 1 & \omega & \omega^2 \end{vmatrix} \]

where \( \omega \) is a cube root of unity, meaning \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

### Step 1: Simplifying the Determinant
First, we simplify the terms \( \omega^4 \) and \( \omega^5 \):
- Since \( \omega^3 = 1 \), we have:
- \( \omega^4 = \omega \)
- \( \omega^5 = \omega^2 \)

Thus, we can rewrite the determinant as:

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

### Step 2: Expanding the Determinant
We can expand this determinant using the first row:

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

### Step 3: Calculating the 2x2 Determinants
Now we compute each of the 2x2 determinants:

1. **First determinant:**
\[ \begin{vmatrix} 1 & 1 + \omega^2 \\ \omega & \omega^2 \end{vmatrix} = 1 \cdot \omega^2 - \omega(1 + \omega^2) = \omega^2 - \omega - \omega^3 = \omega^2 - \omega - 1 \]

2. **Second determinant:**
\[ \begin{vmatrix} \omega & 1 + \omega^2 \\ 1 & \omega^2 \end{vmatrix} = \omega \cdot \omega^2 - 1(1 + \omega^2) = \omega^3 - (1 + \omega^2) = 1 - 1 - \omega^2 = -\omega^2 \]

3. **Third determinant:**
\[ \begin{vmatrix} \omega & 1 \\ 1 & \omega \end{vmatrix} = \omega^2 - 1 \]

### Step 4: Substituting Back
Now substituting back into the expression for \( \Delta \):

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

### Step 5: Simplifying \( \Delta \)
Now simplifying \( \Delta \):

1. The first term:
\[ \omega^2 - \omega - 1 \]

2. The second term:
\[ + \omega^4 = \omega \]

3. The third term:
\[ (1 - \omega)(\omega^2 - 1) = (1 - \omega)(\omega^2 - 1) = \omega^2 - \omega^3 - \omega + \omega^2 = 2\omega^2 - \omega - 1 \]

Combining all these gives us:

\[ \Delta = \omega^2 - \omega - 1 + \omega + 2\omega^2 - \omega - 1 = 3\omega^2 - 2 \]

### Step 6: Finding the Real Part
Now we need to find \( |Re(\Delta)| \). The real part of \( \omega \) is \( -\frac{1}{2} \) and the real part of \( \omega^2 \) is also \( -\frac{1}{2} \).

So,

\[ Re(\Delta) = 3 \left(-\frac{1}{2}\right) - 2 = -\frac{3}{2} - 2 = -\frac{3}{2} - \frac{4}{2} = -\frac{7}{2} \]

Thus,

\[ |Re(\Delta)| = \frac{7}{2} \]

### Final Answer
The final answer is:

\[ |Re(\Delta)| = \frac{7}{2} \]