If \(n\) is an integer, then \((n^3 - n)\) is always divisible by: যদি \(n\) একটি পূর্ণসংখ্যা হয়, তবে \((n^3 - n)\) সর্বদা কোন সংখ্যায় বিভাজ্য?

\[ n^3 - n \]

Factorizing:

\[ n^3 - n = n(n^2 - 1) \]

\[ = n(n-1)(n+1) \]

This is the product of three consecutive integers.

Among any three consecutive integers:

• One number is always divisible by \(3\)
• At least one number is even (divisible by \(2\))

Therefore the product is always divisible by:

\[ 2 \times 3 = 6 \]

Correct Option: C