If n is an integer, then (n3 - n) is always divisible by :

Answer: Option C

$$\begin{array}{rl}& \left(n^3-n\right)\text{ and n is any integer}\\ & \text{put n = 2 so, }{\text{2}}^3-2=6\\ & \text{It will be always divisible by 6}\\ & \text{(put n = 2, 3, 4}....\text{)}\\ & \text{(n = 2, 3, 4}.....\text{)}\end{array}$$