Question

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

Answer 1

Answer :-

step-by-step explanation :-

n³ - n = n (n² - 1) = n (n - 1) (n + 1)

Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.

∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.

If n = 3p, then n is divisible by 3.

If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.

If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.

⇒ n (n – 1) (n + 1) is divisible by 3.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

∴ n = 2q or 2q + 1, where q is some integer.

If n = 2q, then n is divisible by 2.

If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

⇒ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

$\therefore$ n (n-1) (n+1) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

Hence, it is proved.

Answer 2

Solution :-

Let n³ - n be a.

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.