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Home / Elementary Number Theory / The cube of any integer has one of the form 9k, 9k + 1 or, 9k + 8

The cube of any integer has one of the form 9k, 9k + 1 or, 9k + 8

Solution: Let a is any integer. According to division algorithm, every a is of the form 3q, 3q + 1, 3q + 2 i.e., a = 3q or 3q + 1, or, 3q + 2

For,        a = 3q

⟹ a3 = (3q)3

= 27q3

= 3(9q3)

Let, 9q3 = k

∴ a3 = 3k

Again, for a = 3q + 1

⟹ a3 = (3q +1)3

                         = 27q3 + 27q2 + 9q + 1

                         = 9 (3q3 + 3q2 + q) + 1

Let, 3q3 + 3q2 + q = k

∴ a3 = 9k + 1

And for a = 3q + 2

⟹ a3 = (3q + 2)3

= 27q3 + 54q2 + 36q + 8

= 9(3q3 + 6q2 + 4q) + 8

Let, 3q3 + 6q2 + 4q = k

∴ a3 = 9k + 8

So, the cube of any integer has one of the form 9k, 9k + 1 or, 9k + 8. (Proved)

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