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

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Let us consider a and b where a be any positive number and b is equal to 3.

According to Euclid’s Division Lemma

a = bq + r

where r is greater than or equal to zero and less than b (0 ≤ r < b)

a = 3q + r

so r is an integer greater than or equal to 0 and less than 3.

Hence r can be either 0, 1 or 2.

Case 1: When r = 0, the equation becomes

a = 3q

Cubing both the sides

$a^3 = (3q)^3$

$a^3 = 27 q^3$

$a^3 = 9 (3q^3)$

$a^3 = 9m$

where m = $3q^3$

Case 2: When r = 1, the equation becomes

a = 3q + 1

Cubing both the sides

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

$a^3 = (3q)^3 + 13 + 3 × 3q × 1(3q + 1)$

$a3^ = 27q^3 + 1 + 9q × (3q + 1)$

$a^3 = 27q^3 + 1 + 27q^2 + 9q$

$a^3 = 27q^3 + 27q^2 + 9q + 1$

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

$a^3 = 9m + 1$

Where m = $( 3q^3 + 3q^2 + q)$

Case 3: When r = 2, the equation becomes

a = 3q + 2

Cubing both the sides

$a^3 = (3q + 2)^3$

$a^3 = (3q)^3 + 2^3 + 3 × 3q × 2 (3q + 1)$

$a^3 = 27q^3 + 8 + 54q^2 + 36q$

$a^3 = 27q^3 + 54q^2 + 36q + 8$

$a^3 = 9 (3q^3 + 6q^2 + 4q) + 8$

$a^3 = 9m + 8$

Where m = $(3q^3 + 6q^2 + 4q)$therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.

Step-by-step explanation:

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Answer 2

Answer:

refer to the given attachment..

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