Question

prove that for any positive integer a ,its cube is in the form 9m+1

Answer 1

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Let a be any positive integer and b = 3

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

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

Let a be any positive integer and b = 3a = 3q + r, where q ≥ 0 and 0 ≤ r < 3 Therefore, every number can be represented as these three forms. There are three cases.

Let a be any positive integer and b = 3a = 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,

Let a be any positive integer and b = 3a = 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,

Let a be any positive integer and b = 3a = 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 =

Let a be any positive integer and b = 3a = 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,

Let a be any positive integer and b = 3a = 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

Let a be any positive integer and b = 3a = 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

Let a be any positive integer and b = 3a = 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

Let a be any positive integer and b = 3a = 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) + 1a 3 = 9m + 1

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Answer 2

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Let When

$a = 3q + 1$

$So$

$a = {(3q+1)}^3$

$a = 27{q}^3 + 27{q}^2 + 3q + 1$

$a = 9 ( 3{q}^3 + 3{q}^2 + q ) + 1$

$As$

$a = 9m + 1$

$Where$

$m = 3{q}^3 + 3{q}^2 + q$