Question

for any positive integer n, prove that n^3 - n is divisible by 6​

Answer 1

Answer:

n³ - n = n(n² - 1) = n(n - 1) (n+1)

= (n - 1) n(n + 1)

product of 3 consequetive positive integers.

if a = 3q

a(a + 1) (a + 2) = 3q (3q + 1) (3q + 2)

= 3q(2m)

:. 3q + 1, 3q + 2 are two consecutive Integers and their products is also even.

= (6q) m [divisible by 6]

if a = 3q + 1

a(a + 1) (a + 2) = (3q + 1) (3q + 2) (3q + 3)

= [2m] 3(q + 1)

[:. (3q + 1) (3q + 2) = 2m]

= 6m(q + 1) [divisible by 6]

if a = 3q + 2

a(a + 1) (a + 2) = (3q + 2) (3q + 3) (3q + 4)

= (3q + 2) 3(q + 1) (3q + 4)

= multiple of 6

= 6m [divisible by 6]

Step-by-step explanation:

Hope this helps you :)