Let $$a$$ and $$b$$ be two positive integers, and $$a > b$$
$$a = (b \times q) + r$$ where $$q$$ and $$r$$ are positive integers and
$$0 \le r< b$$
Let $$b = 3$$ (If 9 is multiplied by 3 a perfect cube number is obtained)
$$a = 3q + r$$ where $$0\le r < 3$$
(i) if $$r= 0, a = 3q$$ (ii) if $$r = 1, a = 3q + $$1 (iii) if $$r = 2, a = 3q + 2$$
Consider, cubes of these
Case (i) $$a = 3q$$
$$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$$ where $$m=3q^3$$ and '$$m$$' is an integer.
Case (ii) $$a = 3q + 1$$
$$a^3 = (3q + 1)^3$$ $$[(a + b)^3 = a^3+b^3 + 3a^2b+ 3ab^2]$$
$$=27q^3+ 1 + 27q^2 + 9q= 27q^3+ 27q^2+ 9q + 1$$
$$ = 9(3q^3+ 3q^2+ q) + 1 = 9m +1$$
where $$m=3q^3 + 3q^2 + q$$ and '$$m$$' is an integer.
Case (iii) $$a = 3q + 2$$
$$a^3= (3q + 2)^3= 27q^3+ 8 + 54q^2+ 36q$$
$$= 27q^3- 54q^2+ 36q + 8 = 9 (3q^3+ 6q^2 + 4q) + 8$$
$$9m+8$$, where $$m=3q^3+6q^2+4q$$ and $$m$$ is an integer.
$$\therefore$$ cube of any positive integer is either of the form $$9m, 9m+1$$ or $$9m+8$$ for some integer m.