Question

For any positive integer r,prove that n to the power 3
-n is divisible by 6

Answer 1

hope this help..........

Answer 2

Hi Mate !!


Here's the Solution of ur query :-

Let n be any positive integer Which when divided by 6 gives q as quotient and r as remainder.

By Euclid's Division lemma

a = bq + r

where, 0 ≤ r < b

So,

n = 6q + r

where , r = 0 , 1 , 2 , 3 , 4 , 5

____________________

n = 6q

n = 6q + 1

n = 6q + 2

n = 6q + 3

n = 6q + 4

n = 6q + 5


____________

• CASE - 1

n = 6q

n³ - n = ( 6q )³ - 6q

= 216q³ - 6q

= 6 ( 36q³ - q )

= 6m [ Where m = 36q³ - q )

Therefore , it is divisible by 6

________________________

• CASE - 2

n = 6q + 1

n³ - n = ( 6q + 1 )³ - ( 6q + 1 )

{ Using identity :- ( a + b )³ = a³ + b³ + 3a²b + 3ab² }

= 216q³ + 1 + 108q² + 18q - 6q - 1

= 6 ( 36q³ + 18q² + 2q )

= 6m [ Where m = 36q³ + 18q² + q ]

Therefore , it is divisible by 6 .
_____________________


• CASE - 3

n = 6q + 2

n³ - n = ( 6q + 2 )³ - ( 6q + 2 )

= 216q³ + 8 + 216q² + 72q - 6q - 2

= 216q³ + 216q² + 66q + 6

= 6 ( 36q³ + 36q² + 11q + 1 )

= 6m [ Where m = 36q³ + 36q² + 11q + 1 ]

Therefore, It is divisible by 6

_________________


• CASE - 4

n = 6q + 3

n³ - n = ( 6q + 3 )³ - ( 6q + 3 )

= 216q³ + 27 + 324q² + 162q - 6q - 3

= 216q³ + 324q² + 156q + 24

= 6 ( 36q³ + 54q² + 26q + 4 )

= 6m [ Where m = 36q³ + 54q² + 26q + 4 ]

Therefore , It is divisible by 6.

_________________________

• CASE - 5

n = 6q + 4

n³ - n = ( 6q + 4 )³ - ( 6q + 4 )

= 216q³ + 64 + 432q² + 288q - 6q - 4

= 216q³ + 432q² + 282q + 60

= 6 ( 36q³ + 72q² + 47q + 10 )

= 6m [ Where m = 36q³ + 72q² + 47q + 10 ]

Therefore , It is divisible by 6

_____________________

• CASE - 6

n = 6q + 5

n³ - n = ( 6q + 5 )³ - ( 6q + 5 )

= 216q³ + 125 + 540q² + 450q - 6q - 5

= 216q³ + 540q² + 444q + 120

= 6 ( 36q³ + 90q² + 74q + 20 )

= 6m [ Where m = 36q³ + 90q² + 74q + 20 ]

Therefore, It is divisible by 6

__________________

Hence , n³ - n will be divisible by 6 for any positive integer !!