Question

Show that the cube of any positive integer is of the form 9m , 9m+1 or 9m+8;By euclid's division lemma.

Answer 1

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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Answer 2

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,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + 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 is an integer such that m = (3q³ + 6q² + 4q)

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

Hence, it is proved .

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