Let `omega!=1` be a cube root of unity, and `Delta=|(2,2omega,-omega^(2)),(1,1,1),(1,-1,0)|` then `2cos (Delta)=`_____________

Answer

To solve the problem, we need to calculate the determinant \( \Delta = |(2, 2\omega, -\omega^2), (1, 1, 1), (1, -1, 0)| \) where \( \omega \) is a cube root of unity and \( \omega \neq 1 \).

### Step 1: Write the determinant

The determinant can be expressed as follows:

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

### Step 2: Calculate the determinant using the formula

We can calculate the determinant using the formula for a 3x3 matrix:

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

where the matrix is:

\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

Here, \( a = 2, b = 2\omega, c = -\omega^2, d = 1, e = 1, f = 1, g = 1, h = -1, i = 0 \).

Calculating the determinant:

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

### Step 3: Simplify each term

Calculating each term:

1. First term: \( 2(0 + 1) = 2 \)
2. Second term: \( -2\omega(0 - 1) = 2\omega \)
3. Third term: \( -\omega^2(-1 - 1) = 2\omega^2 \)

Putting it all together:

\[ \Delta = 2 + 2\omega + 2\omega^2 \]

### Step 4: Use properties of cube roots of unity

Since \( \omega \) is a cube root of unity, we know that:

\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]

Thus, \( \omega + \omega^2 = -1 \).

Substituting this into our expression for \( \Delta \):

\[ \Delta = 2 + 2(-1) = 2 - 2 = 0 \]

### Step 5: Calculate \( 2\cos(\Delta) \)

Since \( \Delta = 0 \):

\[ 2\cos(\Delta) = 2\cos(0) = 2 \cdot 1 = 2 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{2} \]