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  • For any positive integer n, prove that npower 3 – n is divisible by 6.

For any positive integer n, prove that n3 – n is divisible by 6.

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Let three consecutive positive integers be, x, x + 1 and x + 2.
Divisibility by 3: Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2
.\thereforen = 3p or 3p + 1 or 3p + 2, where p is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 3.
\Rightarrow x (x + 1) (x + 2) is divisible by 3.

Divisibility by 2: Whenever a number is divided 2, the remainder obtained is 0 or 1.
\therefore  n = 2q or 2q + 1, where q is some integer.
So, we can say that one of the numbers among x, x + 1 and x + 2 is always divisible by 2.\Rightarrow
x (x + 1) (x + 2) is divisible by 2.
Since, x (x + 1) (x + 2) is divisible by 2 and 3.
\thereforex (x + 1) (x + 2) is divisible by 6.
Now the given number is:
P = n3 – n
P = n (n2 – 1)
P = n (n – 1) (n + 1)
P = (n – 1) n (n + 1)
Therefore, P is the product of three consecutive numbers.
Now P can be written as:
P = x (x + 1) (x + 2) which is is divisible by 6.
Hence P is divisible by 6.
Hence proved

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