If a is a natural number, then the largest number dividing `(a^(3)-a)` is

Options

A. 4
B. 5
C. 6
D. 7

Answer

To find the largest number that divides \( a^3 - a \) for any natural number \( a \), we can follow these steps:

### Step 1: Factor the expression
We start with the expression \( a^3 - a \). We can factor it as follows:
\[ a^3 - a = a(a^2 - 1) = a(a - 1)(a + 1) \]
This shows that \( a^3 - a \) is the product of three consecutive integers: \( a \), \( a - 1 \), and \( a + 1 \).

### Step 2: Analyze the product of three consecutive integers
The product of three consecutive integers is always divisible by 2 and 3. This is because:
- Among any three consecutive integers, at least one of them is even (hence divisible by 2).
- Among any three consecutive integers, at least one of them is divisible by 3.

### Step 3: Determine the divisibility by 6
Since the product \( a(a - 1)(a + 1) \) is divisible by both 2 and 3, it follows that it is divisible by:
\[ 2 \times 3 = 6 \]
Thus, \( a^3 - a \) is divisible by 6 for any natural number \( a \).

### Step 4: Check for larger divisors
Next, we need to check if there is a larger number than 6 that divides \( a^3 - a \) for all natural numbers \( a \). We can test a few values of \( a \):
- For \( a = 1 \): \( 1^3 - 1 = 0 \) (divisible by any number)
- For \( a = 2 \): \( 2^3 - 2 = 6 \) (divisible by 6)
- For \( a = 3 \): \( 3^3 - 3 = 24 \) (divisible by 6)
- For \( a = 4 \): \( 4^3 - 4 = 60 \) (divisible by 6)

Now, we check the largest common divisor of these results:
- The greatest common divisor of 6, 24, and 60 is 6.

### Conclusion
Therefore, the largest number that divides \( a^3 - a \) for any natural number \( a \) is:
\[ \boxed{6} \]