If `omega` is the imaginary cube root of unity, then what is `(2-omega+2omega^(2))^(27)` equal to ?

Options

A. `3^(27)omega`
B. `-3^(27)omega^(2)`
C. `3^(27)`
D. `-3^(27)`

Explanation

To solve the problem \( (2 - \omega + 2\omega^2)^{27} \) where \( \omega \) is the imaginary cube root of unity, we can follow these steps:

### Step 1: Understand the properties of cube roots of unity
The cube roots of unity are \( 1, \omega, \) and \( \omega^2 \), where:
- \( \omega = e^{2\pi i / 3} \)
- \( \omega^2 = e^{-2\pi i / 3} \)
- The key property is that \( 1 + \omega + \omega^2 = 0 \) and \( \omega^3 = 1 \).

### Step 2: Simplify the expression inside the parentheses
We start with:
\[ 2 - \omega + 2\omega^2 \]
Using the identity \( 1 + \omega + \omega^2 = 0 \), we can express \( \omega^2 \) in terms of \( \omega \):
\[ \omega^2 = -1 - \omega \]
Substituting this into the expression gives:
\[ 2 - \omega + 2(-1 - \omega) = 2 - \omega - 2 - 2\omega = -3\omega \]

### Step 3: Raise the simplified expression to the power of 27
Now we have:
\[ (-3\omega)^{27} \]
Using the property of exponents:
\[ (-3)^{27} \cdot \omega^{27} \]

### Step 4: Simplify \( \omega^{27} \)
Since \( \omega^3 = 1 \), we can reduce the exponent modulo 3:
\[ 27 \mod 3 = 0 \quad \Rightarrow \quad \omega^{27} = (\omega^3)^9 = 1^9 = 1 \]

### Step 5: Combine the results
Thus, we have:
\[ (-3)^{27} \cdot 1 = -3^{27} \]

### Final Answer
The final result is:
\[ (2 - \omega + 2\omega^2)^{27} = -3^{27} \]

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