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