**Answer:**

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented as these three forms.

There are three cases.

**Case 1:** When a = 3q,

a^{3} = (3q)^{3} = 27q^{3} = 9(3q^{3})= 9m

Where m is an integer such that m = 3q^{3}

**Case 2:** When a = 3q + 1,

a^{3} = (3q +1)^{3}

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

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

a^{3} = 9m + 1

Where m is an integer such that m = (3q^{3} + 3q^{2} + q)

**Case 3:** When a = 3q + 2,

a^{3} = (3q +2)^{3}

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

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

a^{3} = 9m + 8

Where m is an integer such that m = (3q^{3} + 6q^{2} + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.