Question

For any positive integar n,prove that n^3-n is divisible by 6

Answer 1

Step-by-step explanation:

2 [ ∵ if P = ab + r , then 0 ≤ r < a by Euclid lemma ]

∴ Let n = 3r , 3r +1 , 3r + 2 , where r is an integer

Case 1 :- when n = 3r

Then, n³ - n is divisible by 3 [∵n³ - n = n(n-1)(n+1) = 3r(3r-1)(3r+1) , clearly shown it is divisible by 3 ]

Case2 :- when n = 3r + 1

e.g., n - 1 = 3r +1 - 1 = 3r

Then, n³ - n = (3r + 1)(3r)(3r + 2) , it is divisible by 3

Case 3:- when n = 3r - 1

e.g., n + 1 = 3r - 1 + 1 = 3r

Then, n³ - n = (3r -1)(3r -2)(3r) , it is divisible by 3

From above explanation we observed n³ - n is divisible by 3 , where n is any positive integers