Question

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

Answer 1

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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...

Answer 2

Step-by-step explanation:

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4)

Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8