Question

use euclid division lemma to show that the cube of any positive integer of the form 9m or 9m+1 or 9m+2​

Answer 1

Solution-----> Let a be any positive integer and

a = 3q + r , where 0 ≤ r ≤ 3

So r can take values 0 , 1 , 2

r = 0 , 1 , 2

Now putting r = 0 , we get ,

a = 3q + 0

=> a = 3q

taking cube of both sides ,

=> a³ = ( 3q )³

=> a³ = 27q³

=> a³ = 9 ( 3q³ )

Let m = 3q³ , putting it we get,

=> a³ = 9 m

Now putting r = 1 , we get ,

a = 3q + 1

Taking cube of both sides

=> a³ = ( 3q + 1 )³

=> a³ = ( 3q )³ + ( 1 )³ + 3 ( 3q ) ( 1 ) ( 3q + 1 )

=> a³ = 27q³ + 1 + 9q ( 3q + 1 )

=> a³ = 27q³ + 1 + 27q² + 9q

=> a³ = 27q³ + 27q² + 9q + 1

=> a³ = 9 ( 3q³ + 3q² + q ) + 1

=> a³ = 9 m + 1 , where m = ( 3q³ + 3q² + q )

Now , putting r = 2 , we get ,

a = 3q + 2

Taking cube both sides we get ,

=> a³ = ( 3q + 2 )³

=> a³ = ( 3q )³ + ( 2 )³ + 3 ( 3q ) ( 2 ) ( 3q + 2 )

=> a³ = 27q³ + 8 + 18q ( 3q + 2 )

=> a³ = 27q³ + 8 + 54q² + 36q

=> a³ = 27q³ + 54q² + 36q + 8

=> a³ = 9 ( 3q³ + 6q² + 4q ) + 8

=> a³ = 9m + 8 , where m = ( 3q³ + 6q² + 4q )

So cube of any positive integer is of the form 9m or ( 9m + 1 ) or ( 9m + 8 )