Question

Show that the cube of any positive integer is of from 9m or 9m + 1 or 9m +8, where m is an integer​

Answer 1

Answer:

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Let's consider a and b where a can be any positive number and b will be equal to 3.

Now , 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 will be 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, then equation becomes

a = 3q

Now , Cubing both the sides

a³ = (3q)³

a3 = 27 q³

a³ = 9 (3q³)

a³ = 9m

where m = 3q³

Case 2:

When r = 1, then equation becomes

a = 3q + 1

Now , Cubing both the sides

a³= (3q + 1)³

a³= (3q)³ + 13 + 3 × 3q × 1(3q + 1)

a³ = 27q³ + 1 + 9q × (3q + 1)

a³ = 27q³ + 1 + 27q² + 9q

a³ = 27q³+ 27q² + 9q + 1

a³= 9 ( 3q³ + 3q² + q) + 1

a³ = 9m + 1

Where m = ( 3q³ + 3q² + q)

Case 3:

When r = 2, then equation becomes

a = 3q + 2

Now , Cubing both the sides

a³ = (3q + 2)³

a³= (3q)³ + 23 + 3 × 3q × 2 (3q + 1)

a³ = 27q³+ 8 + 54q² + 36q

a³ = 27q³+ 54q² + 36q + 8

a³= 9 (3q³+ 6q² + 4q) + 8

a³ = 9m + 8

Where m = (3q³ + 6q² + 4q)

therefore a can be any of the form 9m or 9m + 1 or, 9m + 8

Answer 2

Answer:

Let's consider a and b where a can be any positive number and b will be equal to 3.

Now , 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 will be 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, then equation becomes

a = 3q

Now , Cubing both the sides

a³ = (3q)³

a3 = 27 q³

a³ = 9 (3q³)

a³ = 9m

where m = 3q³

Case 2:

When r = 1, then equation becomes

a = 3q + 1

Now , Cubing both the sides

a³= (3q + 1)³

a³= (3q)³ + 13 + 3 × 3q × 1(3q + 1)

a³ = 27q³ + 1 + 9q × (3q + 1)

a³ = 27q³ + 1 + 27q² + 9q

a³ = 27q³+ 27q² + 9q + 1

a³= 9 ( 3q³ + 3q² + q) + 1

a³ = 9m + 1

Where m = ( 3q³ + 3q² + q)

Case 3:

When r = 2, then equation becomes

a = 3q + 2

Now , Cubing both the sides

a³ = (3q + 2)³

a³= (3q)³ + 23 + 3 × 3q × 2 (3q + 1)

a³ = 27q³+ 8 + 54q² + 36q

a³ = 27q³+ 54q² + 36q + 8

a³= 9 (3q³+ 6q² + 4q) + 8

a³ = 9m + 8

Where m = (3q³ + 6q² + 4q)

therefore a can be any of the form 9m or 9m + 1 or, 9m + 8