**Solution:**

6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5

Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5.

Taking cube of each term, we have,

(6q)^{3} = 216 q^{3} = \(6\times36q^3\) + 0

= 6m + 0, where m is an integer

(6q+1)^{3} = 216q^{3} + 108q^{2} + 18q + 1

= 6(36q^{3} + 18q^{2} + 3q) + 1

= 6m + 1, where m is an integer

(6q+2)^{3} = 216q^{3} + 216q^{2} + 72q + 8

= 6(36q^{3} + 36q^{2} + 12q + 1) +2

= 6m + 2, where m is an integer

(6q+3)^{3} = 216q^{3} + 324q^{2} + 162q + 27

= 6(36q^{3} + 54q^{2} + 27q + 4) + 3

= 6m + 3, where m is an integer

(6q+4)^{3} = 216q^{3} + 432q^{2} + 288q + 64

= 6(36q^{3} + 72q^{2} + 48q + 10) + 4

= 6m + 4, where m is an integer

(6q+5)^{3} = 216q^{3} + 540q^{2} + 450q + 125

= 6(36q^{3} + 90q^{2} + 75q + 20) + 5

= 6m + 5, where m is an integer

**Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.**