For any positive integer 'n' prove that 'n^3-n' is divisible by 6.

not always true: 
n=1 
n^3-n=0 
Maybe you mean if n is 2 or greater? 
 

now for 
 
this product will include 3 consecutive numbers and that means at least one of them will be even and exactly one will be a multiple of 3. Therefore each will be a multiple of 6. 


:)