Question

use eulids division lemme to show that the cube of any positive interger can be either of the form 9m,9m+1,9m+8 where m is any positive intergs​

Answer 1

Answer:

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Step-by-step explanation:

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 =  (3q)3 = 27q3

9(3q3) = 9m

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.

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