Use Euclid's division lemma to show that the cube of any positive integer is of the form $9m, 9m+1$ or $9m+8$.

(N/A) Let $a$ be any positive integer and $b=3$.
By Euclid's division lemma,$a = 3q + r$,where $q \geq 0$ and $0 \leq r < 3$.
Therefore,$a$ can be $3q, 3q+1$,or $3q+2$.
Case $1$: If $a = 3q$,then $a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$,where $m = 3q^3$.
Case $2$: If $a = 3q+1$,then $a^3 = (3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1$,where $m = 3q^3 + 3q^2 + q$.
Case $3$: If $a = 3q+2$,then $a^3 = (3q+2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8$,where $m = 3q^3 + 6q^2 + 4q$.
Thus,the cube of any positive integer is of the form $9m, 9m+1$,or $9m+8$.