If `omega(!=1)` is a cube root of unity, and `(1+omega)^7= A + B omega` . Then (A, B) equals

Options

A. `(-1,1)`
B. `(0,1)`
C. `(1,1)`
D. `(1,0)`

Explanation

To solve the problem, we need to evaluate \( (1 + \omega)^7 \) where \( \omega \) is a cube root of unity, and express it in the form \( A + B\omega \).

### Step-by-Step Solution:

1. **Understanding Cube Roots of Unity**:
The cube roots of unity are the solutions to the equation \( x^3 = 1 \). The roots are \( 1, \omega, \) and \( \omega^2 \), where \( \omega \neq 1 \) and satisfies the equations:
\[ \omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0 \]
From the second equation, we can express \( \omega^2 \) as:
\[ \omega^2 = -1 - \omega \]

2. **Expressing \( (1 + \omega)^7 \)**:
We can rewrite \( (1 + \omega) \) using the identity for cube roots of unity. Since \( \omega^2 = -1 - \omega \), we have:
\[ 1 + \omega = -\omega^2 \]
Therefore,
\[ (1 + \omega)^7 = (-\omega^2)^7 = -(\omega^2)^7 \]

3. **Calculating \( (\omega^2)^7 \)**:
We know that \( \omega^3 = 1 \), so we can reduce the exponent modulo 3:
\[ 7 \mod 3 = 1 \]
Thus,
\[ (\omega^2)^7 = \omega^{14} = \omega^{3 \cdot 4 + 2} = (\omega^3)^4 \cdot \omega^2 = 1^4 \cdot \omega^2 = \omega^2 \]
Therefore,
\[ (1 + \omega)^7 = -\omega^2 \]

4. **Expressing \( -\omega^2 \) in terms of \( A + B\omega \)**:
We can express \( -\omega^2 \) using the relation \( \omega^2 = -1 - \omega \):
\[ -\omega^2 = -(-1 - \omega) = 1 + \omega \]
Thus, we can write:
\[ (1 + \omega)^7 = 1 + \omega \]

5. **Identifying \( A \) and \( B \)**:
From the expression \( A + B\omega = 1 + \omega \), we can see that:
\[ A = 1 \quad \text{and} \quad B = 1 \]

### Final Answer:
Thus, the values of \( (A, B) \) are:
\[ (A, B) = (1, 1) \]