Question

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

Answer 1

n^3 - n = n(n^2 - 1)

            n^2 - 1 = (n-1)(n+1)

Therefore, n^3 - n = n(n-1)(n+1)

It is the product of three consecutive numbers in which there will be a multiple of 2 and a multiple of 3. For example, take 4*5*6, here 4 is multiple of 2 and 6 is multiple of 3, which will also be a multiple of 6 (2*3) .

Therefore in any value of n , n^3 - n will be divisible by 6 .

Hope it helps

Answer 2

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.

∴ 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 solved ✅✅.