Question

show that the cube of any positive integers of the form 9m,9m+1 or 9m+8 for some integer m​

Answer 1

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Answer 2

We know that, any number can be expressed as a form of

a = bq + r

where, 0 ≤ r < b

So, if b is 3, then any number can be expressed as 3q, 3q + 1 or 3q + 2 for some positive integer q.

So, now if the number is 3q, then cube = (3q)³ = 27q³

Now, 27q³ = 9(3q³)

Now, let 3q³ = m for some integer m

so, 27q³ = 9m

Now if number is of the form of 3q + 1, then cube =

(3q + 1)³ = 27q³ + 1 + 3(3q)(1)(3q + 1)

⇒ 27q³ + 9q² + 9q + 1

⇒ 9(3q³ + q² + q) + 1

Now, let 3q³ + q² + q = m for some integer m

⇒ 9m + 1

Now if the number is 3q + 2, then the cube of the number would be

(3q + 2)³ = 27q³ + 8 + 3(3q)(2)(3q + 2)

⇒ 27q³ + 54q² + 36q + 8

⇒ 9(3q³ + 6q² + 4q) + 8

Now, let 3q³ + 6q² + 4q = m for some integer m

⇒ 9m + 8

Hence,cube of any positive integer can be of the form of 9m, 9m + 1, 9m + 8