If `omega(ne 1)` be a cube root of unity and `(1+omega)^(7)=A+Bomega`, then A and B are respectively the numbers.

Options

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

Answer

To solve the problem, we need to find the values of \( A \) and \( B \) in the expression \( (1 + \omega)^7 = A + B\omega \), where \( \omega \) is a cube root of unity and \( \omega \neq 1 \).

**Step 1: Understand the properties of cube roots of unity.**
The cube roots of unity are given by the equation \( x^3 = 1 \). The roots are:
- \( 1 \)
- \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \)
- \( \omega^2 = e^{4\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \)

These roots satisfy the following properties:
1. \( 1 + \omega + \omega^2 = 0 \)
2. \( \omega^3 = 1 \)

From the first property, we can express \( 1 + \omega \) as:
\[ 1 + \omega = -\omega^2 \]

**Step 2: Substitute \( 1 + \omega \) in the expression.**
Now, we can rewrite \( (1 + \omega)^7 \):
\[ (1 + \omega)^7 = (-\omega^2)^7 = -(\omega^2)^7 = -\omega^{14} \]

**Step 3: Simplify \( \omega^{14} \).**
Since \( \omega^3 = 1 \), we can reduce the exponent modulo 3:
\[ 14 \mod 3 = 2 \quad \text{(since \( 14 = 4 \times 3 + 2 \))} \]
Thus,
\[ \omega^{14} = \omega^2 \]
Therefore,
\[ (1 + \omega)^7 = -\omega^2 \]

**Step 4: Express \( -\omega^2 \) in the form \( A + B\omega \).**
We know that \( \omega^2 = -1 - \omega \) (from \( 1 + \omega + \omega^2 = 0 \)):
\[ -\omega^2 = 1 + \omega \]
This means we can write:
\[ (1 + \omega)^7 = 1 + \omega \]

**Step 5: Identify \( A \) and \( B \).**
From the expression \( A + B\omega \), we can see:
- \( A = 1 \)
- \( B = 1 \)

Thus, the values of \( A \) and \( B \) are:
\[ A = 1, \quad B = 1 \]

**Final Answer:**
The values of \( A \) and \( B \) are respectively \( 1 \) and \( 1 \).

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