If'n' be any natural number, then by which largest number `n^3-n` is always divisible ?

Options

A. 3
B. 6
C. 12
D. 18

Answer

To determine the largest number by which \( n^3 - n \) is always divisible for any natural number \( n \), we can follow these steps:

### Step 1: Factor the expression \( n^3 - n \)
We start with the expression:
\[ n^3 - n \]
This can be factored as:
\[ n(n^2 - 1) = n(n - 1)(n + 1) \]

### Step 2: Analyze the factors
The expression \( n(n - 1)(n + 1) \) consists of three consecutive integers: \( n \), \( n - 1 \), and \( n + 1 \).

### Step 3: Determine divisibility by 2 and 3
Among any three consecutive integers, at least one of them is divisible by 2, and at least one of them is divisible by 3. Therefore, the product \( n(n - 1)(n + 1) \) is always divisible by:
- \( 2 \) (from one of the integers),
- \( 3 \) (from one of the integers).

### Step 4: Calculate the least common multiple (LCM)
Since we have established that \( n(n - 1)(n + 1) \) is divisible by both 2 and 3, we can find the least common multiple:
\[ \text{LCM}(2, 3) = 6 \]

### Step 5: Conclusion
Thus, \( n^3 - n \) is always divisible by \( 6 \) for any natural number \( n \).

### Final Answer
The largest number by which \( n^3 - n \) is always divisible is:
\[ \boxed{6} \]