Question

For any positive integer n, prove that n^3 - n is divisible by 6.

Answer 1

n³ - n can be written as :

n ( n² - 1 ) [ By taking n as common ]

Now apply the formula : a² - b² = ( a + b )( a - b )

n ( n² - 1 ) = n ( n + 1 )( n - 1 )

Now every positive integer belongs to { 1 , 2 , 3 ............... }

Let p ( n ) be n ( n + 1 )( n - 1 )

Putting n = 1

p ( 1 ) = 1 × 2 × 3

         = 6

Hence it is divisible by 6 as 6 | p ( 1 ) !

Putting n = 2

p ( 2 ) = 2 × 3 × 4

         = 6 × 4

Hence 6 | p (2)

Now put n = 3

p ( 3 ) = 3 × 4 × 5

         = 3 × 2 × 2 × 5

         = 6 × 10

Hence 6 | p(3)

By mathematical induction for any natural number p ( n ) should also be divisible by 6

Hence 6 divides n³ - n

Hope it helps :-)

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Answer 2

Hey there !!

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.