Question

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

Answer 1

n³ - n

= n(n²-1)

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

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

n³-n is the product of three consecutive numbers. It is obvious that at least one of three consecutive numbers is even and divisible by 2 .

Also, n(n+1)(n-1) is divisible by 3 because

n+n+1+n-1 = 3n ( 3n is divisible by 3 :- Divisibility rule of three ) .

As the result of n³-n is divisible by both 2 ,3 it is divisible by 6 as prime factors of 6 are 2 ,3 .

Hence proved!

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 ✅✅.