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

⟹ a^{3} = (3q)^{3}

= 27q^{3}

= 3(9q^{3})

Let, 9q^{3} = k

∴ a^{3} = 3k

Again, for a = 3q + 1

** ⟹ a ^{3} = (3q +1)^{3}**

** = 27q ^{3} + 27q^{2} + 9q + 1**

** = 9 (3q ^{3} + 3q^{2} + q) + 1**

Let, 3q^{3} + 3q^{2} + q = k

∴ a^{3} = 9k + 1

And for a = 3q + 2

⟹ a^{3} = (3q + 2)^{3}

= 27q^{3} + 54q^{2} + 36q + 8

= 9(3q^{3} + 6q^{2} + 4q) + 8

Let, 3q^{3} + 6q^{2} + 4q = k

∴ a^{3} = 9k + 8