for any positive integer n, prove that n cube minus n is divisible by 6​

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1. Answer:

   When a number is divided by 3, the possible remainders are 0 or 1 or 2.

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

   Case 1: Consider n = 3p

   Then n is divisible by 3.

   Case 2: Consider n = 3p + 1

   Then n – 1 = 3p + 1 –1

   ⇒ n -1 = 3p is divisible by 3.

   Case 3: Consider n = 3p + 2

   Then n + 1 = 3p + 2 + 1

   ⇒ n+1 = 3p + 3

   ⇒ n+1 = 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, when a number is divided by 2, the possible remainders are 0 or 1.

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

   Case 1: Consider n = 2q

   Then n is divisible by 2.

   Case 2: Consider n = 2q + 1

   Then n–1 = 2q + 1 – 1

   n – 1 = 2q is divisible by 2 and

   n + 1 = 2q + 1 + 1

   n +1 = 2q + 2

   n+1= 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, as per the divisibility rule of 6, the given number is divisible by six.

   n^3 – n = n (n – 1) (n + 1) is divisible by 6.

   HOPE IT HELPS YOU

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