Question

use euclid division lemma to show that the cube of any positive integer is of the form 9m 9m+1 or 9m+8

Answer 1

Hey

Here is your answer,

Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,

Where m is an integer such that m =
Case 2: When a = 3q + 1,
a 3 = (3q +1) 3
a 3 = 27q 3 + 27q 2 + 9q + 1
a 3 = 9(3q 3 + 3q 2 + q) + 1
a 3 = 9m + 1
Where m is an integer such that m = (3q 3 + 3q 2 + q)
Case 3: When a = 3q + 2,
a 3 = (3q +2) 3
a 3 = 27q 3 + 54q 2 + 36q + 8
a 3 = 9(3q 3 + 6q 2 + 4q) + 8
a 3 = 9m + 8
Where m is an integer such that m = (3q 3 + 6q 2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Hope it helps you!

Answer 2

Answer:

It is possible.

Step-by-step Explanation:

Let a be any positive integer and b = 3

∵ a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0, 1, 2

Therefore, every number can be represented as these three forms. There are three cases.

→ Case 1: When a = 3q,

⇒ a = ( 3q )³

⇒ a = 9( 3q³ )

⇒ a = 9m [ Where m = 3q³ ]

→ Case 2: When a = 3q + 1,

⇒ a = (3q +1)³

⇒ a = 27q³ + 27q² + 9q + 1

⇒ a = 9(3q³ + 3q² + q) + 1

⇒ a = 9m + 1 [ Where m = 3q³ + 3q² + q ) ]

→ Case 3: When a = 3q + 2,

⇒ a = (3q +2)³

⇒ a = 27q³ + 54q² + 36q + 8

⇒ a = 9(3q³ + 6q² + 4q) + 8

⇒ a = 9m + 8 [ Where m = (3q³ + 6q² + 4q) ]

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Thanks ..!!