Question

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

Answer 1

after that
a=3q+2
cubing both side
a³=(3q+2)³
a³=(3q)³+(2)³+3x3qx2(3q+2)......by identity
a³=27q³+8+54q²+36q
a³=(27q³+54q²+36q)+8
a³=9(3q³+6q²+4q)+8.....where 3q³+6q²+4q=M
a³=9(m)+8
a³=9m +8
hence ,proved

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 ..!!