Question

Prove that the any positive integer and prove that n cube minus and

Answer 1

1. Question⇒  For any positive integer n, prove that n3-n is divisible by 6.

Step-by-step explanation:

• The Sage
• n³ - n = n(n²-1) = n(n -1)(n + 1) is divided by 3 then possible reminder is 0, 1 and 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