**Solution:**

Let a be the positive integer and b = 4.

Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.

(4q)^{3} = 64q^{3} = 4(16q^{3})

= 4m, where m is some integer.

(4q + 1)^{3} = 64q^{3} + 48q^{2} + 12q + 1

= 4(16q^{3} + 12q^{2} + 3) + 1

= 4m + 1, where m is some integer.

(4q + 2)^{3} = 64q^{3} + 96q^{2} + 48q + 8

= 4(16q^{3} + 24q^{2} + 12q + 2)

= 4m, where m is some integer.

(4q + 3)^{3} = 64q^{3} + 144q^{2} + 108q + 27

= 4(16q^{3} + 36q^{2} + 27q + 6) + 3

= 4m + 3, where m is some integer.

**Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.**