# prove n3-n is always divisible by 6

Since  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)

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proof:- if n=1,then

1 to the power 3 - 1=1-1=0 which is divisible by 6.

if n=2,then

2 to the power 3 - 2 = 8-2 = 6 which is divisible by 6.

if n=3,then

3 to the power 3 - 3 = 27-3 = 24 which is divisible by 6

and so on.............. doing this,

it is proved that n to the power 3 is always divisible by 6.

thankyou!

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