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

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for any positive integer $a$ and $6$,there exist non-negative integers $q$ and $r$ such that $a = 6q + r$,where $0 \leq r < 6$.
Taking the cube of both sides:
$a^3 = (6q + r)^3 = 216q^3 + r^3 + 3(6q)(r)(6q + r)$
$a^3 = (216q^3 + 108q^2r + 18qr^2) + r^3 \quad \dots(i)$
Case $I$: If $r = 0$,$a^3 = 216q^3 = 6(36q^3) = 6m$,where $m = 36q^3$.
Case $II$: If $r = 1$,$a^3 = (216q^3 + 108q^2 + 18q) + 1 = 6(36q^3 + 18q^2 + 3q) + 1 = 6m + 1$.
Case $III$: If $r = 2$,$a^3 = (216q^3 + 216q^2 + 72q) + 8 = (216q^3 + 216q^2 + 72q + 6) + 2 = 6(36q^3 + 36q^2 + 12q + 1) + 2 = 6m + 2$.
Case $IV$: If $r = 3$,$a^3 = (216q^3 + 324q^2 + 162q) + 27 = (216q^3 + 324q^2 + 162q + 24) + 3 = 6(36q^3 + 54q^2 + 27q + 4) + 3 = 6m + 3$.
Case $V$: If $r = 4$,$a^3 = (216q^3 + 432q^2 + 288q) + 64 = (216q^3 + 432q^2 + 288q + 60) + 4 = 6(36q^3 + 72q^2 + 48q + 10) + 4 = 6m + 4$.
Case $VI$: If $r = 5$,$a^3 = (216q^3 + 540q^2 + 450q) + 125 = (216q^3 + 540q^2 + 450q + 120) + 5 = 6(36q^3 + 90q^2 + 75q + 20) + 5 = 6m + 5$.
Thus,the cube of any positive integer of the form $6q + r$ is of the form $6m + r$.