If x is an integer, then `(x + 1)^(4) - (x-1)^(4)` is always divisible by

Options

A. 6
B. 8
C. 9
D. 12

Answer

To solve the problem, we need to determine if the expression \((x + 1)^4 - (x - 1)^4\) is always divisible by any of the given options: 6, 8, 9, or 12, where \(x\) is an integer.

### Step-by-Step Solution:

1. **Expand the Expression**:
We start by expanding the expression \((x + 1)^4 - (x - 1)^4\).
Using the binomial theorem, we can expand both terms:
\[ (x + 1)^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 \]
\[ (x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 \]

2. **Subtract the Two Expansions**:
Now, we subtract the second expansion from the first:
\[ (x + 1)^4 - (x - 1)^4 = (x^4 + 4x^3 + 6x^2 + 4x + 1) - (x^4 - 4x^3 + 6x^2 - 4x + 1) \]
Simplifying this gives:
\[ = 4x^3 + 4x + 4x^3 + 4x = 8x^3 + 8x \]

3. **Factor the Expression**:
We can factor out the common term:
\[ 8(x^3 + x) \]

4. **Analyze the Factor**:
The expression \(8(x^3 + x)\) shows that it is clearly divisible by 8, since 8 is a factor of the entire expression.

5. **Check Divisibility by Other Options**:
- **Divisibility by 6**: \(8(x^3 + x)\) is not necessarily divisible by 6 for all integers \(x\).
- **Divisibility by 9**: \(8(x^3 + x)\) is not necessarily divisible by 9 for all integers \(x\).
- **Divisibility by 12**: \(8(x^3 + x)\) is not necessarily divisible by 12 for all integers \(x\).

6. **Conclusion**:
Therefore, the only option that is always a divisor of the expression is **8**.

### Final Answer:
The expression \((x + 1)^4 - (x - 1)^4\) is always divisible by **8**.