Solution

By Euclid's division algorithm,
$$ a = 6q + r $$ ..... (i)
where $$ a , q $$ and $$ r $$ are non-negative integers $$ 0 \leq r < 6 $$ i.e., $$ r = 0 , 1 , 2 , 3 , 4 , 5 $$.
Cubing (i) both sides , we get
$$ (a)^3 = (6q + r)^3 $$
$$ \Rightarrow \,a^3 = (6q)^3 + (r)^3 + 3(6q)^2 (r) + 3(6q) (r)^2 $$
$$ \Rightarrow \,a^3= 6^3q^3 + r^3 + 3 (6)^2q^2r + 6 \times 3qr^2 $$
$$ \Rightarrow \,a^3 = 6[36q^3 + 18q^2r + 3qr^2] + r^3 $$ ....(ii)

When $$ r = 0 $$
$$ a^3 = 6[36q^3 + 18q^2 \times 0 + 3q 0^2] + 0^3 $$ [From (ii)]
$$ \Rightarrow \,a^3= 6[36q^3] $$
$$ \Rightarrow \,a^3 = 6m $$ is perfect cube for some integer $$m$$ such that $$ m = 36q^2 $$

When $$ r = 1 $$
$$a^3 = 6[36q^3 + 18q^2 \times 1 + 3q 1^2] + 1^3 $$
$$\Rightarrow \,a^3 = 6[36q^3 + 18q^2 \times 1 + 3q ] + 1 $$
$$ \Rightarrow \,a^3 = 6m + 1 $$ is perfect cube for some integer $$m$$ such that $$ m = (36q^2 + 18q^2 + 3q) $$

When $$ r = 2 $$
$$a^3 = 6[36q^3 + 18q^2 \times 2 + 3q \times 2^2] + 2^3 $$ [From(ii)]
$$\Rightarrow \,a^3 = 6[36q^3 + 36q^2 + 12q] + 6 + 2 $$
$$ \Rightarrow \,a^3= 6[36q^3 + 36q^2 + 12q + 1] + 2 $$
$$ \Rightarrow \,a^3 = 6m + 2 $$ is perfect cube for some integer $$m$$ such that $$ m = 36q^3 + 36q^2 + 12q + 1 $$

When $$ r = 3 $$
$$a^3 = 6[36q^3 + 18q^2 \times 3 + 3q \times 3^2] + 3^3 $$ [From(ii)]
$$ \Rightarrow \,a^3 = 6[36q^3 + 54q^2 + 27q] + 24 + 3 $$
$$ \Rightarrow \,a^3 = 6[36q^3 + 54q^2 + 27q + 4] + 3 $$
$$ \Rightarrow \,a^3 = 6m + 3 $$
So , $$ (6m + 3) $$ is perfect cube for some integer $$ m $$ such that $$ m = 36q^2 + 54q^2 + 27q + 4 $$

When $$ r = 4 $$
$$a^3 = 6[36q^3 + 18q^2(4) + 3q4^2] + 4^3 $$ [From(ii)]
$$ \Rightarrow \,a^3 = 6 [36q^3 + 72q^2 + 48q] + 60 + 4 $$
$$ \Rightarrow \,a^3 = 6 [36q^3 + 72q^2 + 48q + 10] + 4 $$
$$ \Rightarrow \,a^3 = 6m + 4 $$
So , $$ (6m + 4) $$ is perfect cube for some integer $$m$$ such that $$ m = 36q^3 + 72q^2 + 48q + 10 $$

When $$ r = 5 $$
$$ a^3 = 6[36q^3 + 18q^2(5) + 3q(5)^2] +(5)^3 $$ [From(ii)]
$$ \Rightarrow \,a^3 = 6[36q^3 + 90q^2 + 75q ] + 120 + 5 $$
$$ \Rightarrow \,a^3= 6[36q^3 + 90q^2 75q + 20] + 5 $$
$$ \Rightarrow \,a^3 = 6m + 5 $$
So, $$ ,(6m + 5) $$ is perfect cube for some integer of $$ m = 36q^3 + 90q^2 + 75q + 20 $$

Hence , cubes of positive integers of the form $$(6q+r)$$ is also of the form $$ (6m + r) $$ where $$m$$ is a specified integer and $$ r = 0, 1 , 2 , 3 , 4 , 5 $$.