Question

for any positive integer n prove that n 3 minus n is divisible by 6​

Answer 1

Answer:

Step-by-step explanation:

Let :-

a = n³ - n

  = n(n² - 1)

  = n(n - 1)(n + 1)

  = (n - 1)n (n + 1)

1) Now out of three (n - 1),n and (n + 1) one must be even so a is divisible by 2

2) Also (n - 1),n and (n + 1) are three consecutive integers thus as proved a must be divisible by 3

From (1) and (2)

a must be divisible by 2 × 3 = 6

Hence n³ - n is divisible by 6 for any positive integer n