`(1+w)^7=A+Bw` where w is the imaginary cube root of of a unity and `A, B in R`, find the ordered pair (A, B).

Answer

To solve the problem `(1 + w)^7 = A + Bw` where \( w \) is the imaginary cube root of unity, we will follow these steps:

### Step 1: Understand the properties of \( w \)
The imaginary cube roots of unity are given by:
- \( w = e^{2\pi i / 3} \) (which is \( -\frac{1}{2} + \frac{\sqrt{3}}{2} i \))
- \( w^2 = e^{-2\pi i / 3} \) (which is \( -\frac{1}{2} - \frac{\sqrt{3}}{2} i \))
- The important properties are:
- \( 1 + w + w^2 = 0 \)
- \( w^3 = 1 \)

### Step 2: Rewrite \( 1 + w \)
Using the property \( 1 + w + w^2 = 0 \), we can express \( 1 + w \) as:
\[ 1 + w = -w^2 \]

### Step 3: Raise \( 1 + w \) to the power of 7
Now we need to compute \( (1 + w)^7 \):
\[ (1 + w)^7 = (-w^2)^7 = -w^{14} \]

### Step 4: Simplify \( w^{14} \)
Since \( w^3 = 1 \), we can reduce \( w^{14} \):
\[ w^{14} = w^{3 \cdot 4 + 2} = (w^3)^4 \cdot w^2 = 1^4 \cdot w^2 = w^2 \]
Thus,
\[ (1 + w)^7 = -w^2 \]

### Step 5: Express \( -w^2 \) in terms of \( A + Bw \)
We know that \( w^2 = -1 - w \) (from \( 1 + w + w^2 = 0 \)), so:
\[ -w^2 = -(-1 - w) = 1 + w \]

### Step 6: Compare with \( A + Bw \)
Now we can compare \( 1 + w \) with \( A + Bw \):
\[ A + Bw = 1 + 1w \]
From this, we can see that:
- \( A = 1 \)
- \( B = 1 \)

### Conclusion
The ordered pair \( (A, B) \) is:
\[ \boxed{(1, 1)} \]