Question

Prove that for any positive integer n , n³ - n is divisible by 6

Please give relevant solutions

Answer 1

n^3 -n = n*(n^2-1) = n*(n+1)*(n-1)
[a^2 - b^2 = (a+b)*(a-b) identity]

so, n^3 - n = (n-1)*n*(n+1)
Now we can see that R.H.S represents the product of 3 consecutive numbers.
The product of 3 consecutive positive integers is always divisible by '6'.
Let n =2
1*2*3 = 6.... always divisible by '6'
Let n=3
2*3*4 = 12.... again divisible by '6'
Let n= 100
99*100*101... again divisible by '6'

OR
n^3 - n = (n-1)*n*(n+1)
When we divide any number by '2', the remainder obtained is either 0 or 1.
Number = 2k (rem = 0) or 2k+1(rem = 1)
n = 2k.....divisible by '2'
n = 2k+1...n-1 = 2k+1-1 = 2k..... again divisible by '2'
So, whatever the number be: (n-1)*n*(n+1) is always divisible by '2'

When we divide any number by '3', the remainder is 0 or 1 or 2.
Number = 3k (rem = 0) or 3k+1 (rem=1) or 3k+2(rem=2)
n=3k...always divisible by '3'
n= 3k+1....n-1 = 3k...again divisible by '3'
n= 3k+2...n+1 = 3k+3= 3(k+1)...again divisible by '3'
Hence, (n-1)*n*(n+1) is always divisible by '3'

If a number is always divisible by both 2 & 3, then the number is divisible by '6'
Hence, n^3 - n is always divisible by '6' (for n = positive integer)
Hope it helps.

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)

THANKS

#BeBrainly.