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 simplify the expression \( (x + 1)^4 - (x - 1)^4 \) and determine what it is always divisible by when \( x \) is an integer.

### Step-by-Step Solution:

1. **Recognize the Difference of Squares**:
We can use the identity \( a^2 - b^2 = (a - b)(a + b) \). Here, we can treat \( (x + 1)^4 \) as \( a^2 \) and \( (x - 1)^4 \) as \( b^2 \).

\[ (x + 1)^4 - (x - 1)^4 = [(x + 1)^2 - (x - 1)^2][(x + 1)^2 + (x - 1)^2] \]

2. **Calculate \( (x + 1)^2 - (x - 1)^2 \)**:
Using the difference of squares again:
\[ (x + 1)^2 - (x - 1)^2 = [(x + 1) - (x - 1)][(x + 1) + (x - 1)] \]
Simplifying this:
\[ = [2][2x] = 4x \]

3. **Calculate \( (x + 1)^2 + (x - 1)^2 \)**:
Expanding both squares:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (x - 1)^2 = x^2 - 2x + 1 \]
Adding these:
\[ (x + 1)^2 + (x - 1)^2 = (x^2 + 2x + 1) + (x^2 - 2x + 1) = 2x^2 + 2 \]

4. **Combine the Results**:
Now substituting back into our expression:
\[ (x + 1)^4 - (x - 1)^4 = (4x)(2x^2 + 2) \]
Simplifying:
\[ = 8x(x^2 + 1) \]

5. **Determine Divisibility**:
The expression \( 8x(x^2 + 1) \) is always divisible by 8 since \( 8 \) is a factor.

Additionally, since \( x(x^2 + 1) \) is an integer for any integer \( x \), the entire expression is divisible by \( 8 \).

### Conclusion:
Thus, \( (x + 1)^4 - (x - 1)^4 \) is always divisible by **8**.