If x is any natural number, then `x^(5) -x` is divisible by:

Options

A. 6 but not by 10
B. 10 but not by 6
C. Both 6 and 10
D. Neither 6 nor 10

Answer

To determine if \( x^5 - x \) is divisible by certain numbers when \( x \) is any natural number, we can start by factoring the expression.

### Step 1: Factor the expression
We can factor \( x^5 - x \) as follows:
\[ x^5 - x = x(x^4 - 1) \]
Next, we can factor \( x^4 - 1 \) further:
\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \]
And \( x^2 - 1 \) can be factored as:
\[ x^2 - 1 = (x - 1)(x + 1) \]
So, the complete factorization of \( x^5 - x \) is:
\[ x^5 - x = x(x - 1)(x + 1)(x^2 + 1) \]

### Step 2: Analyze the factors
Now we have:
\[ x^5 - x = x(x - 1)(x + 1)(x^2 + 1) \]
This product consists of four factors: \( x \), \( x - 1 \), \( x + 1 \), and \( x^2 + 1 \).

### Step 3: Check divisibility by 2
Among the factors:
- \( x \) is even if \( x \) is an even natural number.
- \( x - 1 \) is even if \( x \) is odd.
- \( x + 1 \) is even if \( x \) is odd.

Thus, in either case, at least one of the factors \( x \), \( x - 1 \), or \( x + 1 \) is always even. Therefore, \( x^5 - x \) is divisible by 2.

### Step 4: Check divisibility by 3
Among the three consecutive integers \( x - 1 \), \( x \), and \( x + 1 \), at least one of them must be divisible by 3. Therefore, \( x^5 - x \) is divisible by 3.

### Step 5: Check divisibility by 6
Since \( x^5 - x \) is divisible by both 2 and 3, it is also divisible by their product, which is 6.

### Conclusion
Thus, we conclude that \( x^5 - x \) is divisible by 6 for any natural number \( x \).