Question

For any positive integer n, prove that n3 – n is divisible by ❌❌❌​

Answer 1

Step-by-step explanation:

The condition for any number to be divisible by 6 is that the number must be individually divisible by 3 and 2.

Check whether n

3

−n is divisible by 3.

n

3

−n=n(n+1)(n−1)

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

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

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

If n=3p+1, then n−1=3p+1−1=3p. The number is divisible by 3.

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

So, any number in the form of n

3

−n=n(n+1)(n−1) is divisible by 3.

Check whether n

3

−n is divisible by 2.

When a number is divided by 2, the remainder obtained is either 0 or 1 by the remainder theorem.

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

If n=2p, then the number is divisible by 2.

If n=2p+1 then n−1=2p+1−1=2p. The number is divisible by 2.

So, any number in the form of n

3

−n=n(n+1)(n−1) is divisible by 2.

Since, the given number n

3

−n=n(n+1)(n−1) is divisible by both 3 and 2. Therefore, according to the divisibility rule of 6, the given number is divisible by 6.

Hence, n

3

−n=n(n+1)(n−1) is divisible by 6.

Answer 2

Answer:

$\huge \red {solution}$

(n – 1) , n, (n + 1) are consecutive integers thus as proved a must be divisible by 3. Thus, n³ – n is divisible by 6 for any positive integer n.