Let a, b, c be the sides of `Delta`ABC opposite to angles
A, b,C respecitvely.
Let `alpha = sum_(r=0)^(n) ""^(n)C_(r) b^(n-r) c^(r) cos{rB - (n-r)C}`
and `beta = sum_(r=0)^(n) ""^(n)C_(r) b^(n-r) c^(r) sin{rB - (n-r)C}`
Statement -1: `alpha = alpha^(n)`
Statement-2: `beta = alpha^(n)`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to evaluate the expressions for \(\alpha\) and \(\beta\) and analyze the given statements.

### Step-by-Step Solution:

1. **Define the Expressions**:
We have:
\[ \alpha = \sum_{r=0}^{n} \binom{n}{r} b^{n-r} c^{r} \cos(rB - (n-r)C) \]
\[ \beta = \sum_{r=0}^{n} \binom{n}{r} b^{n-r} c^{r} \sin(rB - (n-r)C) \]

2. **Combine \(\alpha\) and \(\beta\)**:
We can express \(\alpha + i\beta\) as:
\[ \alpha + i\beta = \sum_{r=0}^{n} \binom{n}{r} b^{n-r} c^{r} e^{i(rB - (n-r)C)} \]
This uses Euler's formula, where \(e^{ix} = \cos x + i\sin x\).

3. **Factor the Exponential**:
The expression can be factored as:
\[ \alpha + i\beta = \sum_{r=0}^{n} \binom{n}{r} b^{n-r} c^{r} e^{i(rB - (n-r)C)} = (b e^{iC} + c e^{iB})^n \]
This is a result of the binomial expansion.

4. **Evaluate the Magnitude**:
Let \(z = b e^{iC} + c e^{iB}\). Then:
\[ \alpha + i\beta = z^n \]
Hence, we can write:
\[ \alpha = \text{Re}(z^n) \quad \text{and} \quad \beta = \text{Im}(z^n) \]

5. **Analyze the Statements**:
- **Statement 1**: \(\alpha = \alpha^n\)
- **Statement 2**: \(\beta = \alpha^n\)

From the expression for \(\alpha\), we can see that \(\alpha\) is indeed equal to \(\alpha^n\) under certain conditions (specifically when \(\beta = 0\)). Therefore, Statement 1 can be considered true.

For Statement 2, we derived that \(\beta\) can be expressed in terms of \(b\) and \(c\) and it does not equal \(\alpha^n\) in general. Thus, Statement 2 is false.

### Conclusion:
- **Statement 1** is true: \(\alpha = \alpha^n\)
- **Statement 2** is false: \(\beta \neq \alpha^n\)