`(x^(n)-a^(n))` is completely divisible by `(x+a)`, when

Options

A. n is any natural number
B. n is an even natural number
C. n is an odd natural number
D. n is prime

Answer

To determine when \( x^n - a^n \) is completely divisible by \( x + a \), we can analyze the expression based on the value of \( n \).

### Step-by-Step Solution

1. **Understanding the Expression**:
We start with the expression \( x^n - a^n \) and want to check its divisibility by \( x + a \).

2. **Using the Factorization for Even \( n \)**:
- When \( n \) is an even natural number, we can use the difference of squares:
\[ x^n - a^n = (x^{n/2} - a^{n/2})(x^{n/2} + a^{n/2}) \]
- Continuing this process, we can factor it down to:
\[ x^2 - a^2 = (x - a)(x + a) \]
- This shows that \( x + a \) is a factor of \( x^n - a^n \) when \( n \) is even.

3. **Using the Factorization for Odd \( n \)**:
- When \( n \) is an odd natural number, we can use the identity for the difference of cubes:
\[ x^3 - a^3 = (x - a)(x^2 + ax + a^2) \]
- Similarly, for \( n = 1 \):
\[ x^1 - a^1 = x - a \]
- In both cases, \( x + a \) is not a factor, indicating that \( x^n - a^n \) is not divisible by \( x + a \) when \( n \) is odd.

4. **Conclusion**:
- Therefore, we conclude that \( x^n - a^n \) is divisible by \( x + a \) **only when \( n \) is an even natural number**.

### Final Answer
\( x^n - a^n \) is completely divisible by \( x + a \) when \( n \) is an even natural number.