Use euclid division algorithm to show …

Let a be any positive integer and b = 9.
Then by Euclid’s division lemma
a = 9q + r where q ≥0 and 0 ≤ r < 8
So, a = 9q, 9q + 1, 9q + 2, 9q + 3, 9q + 4,
9q + 5, 9q + 6, 9q + 7, 9q + 8
All these numbers can be expressed as
a = 9q = 3(3q) = 3n
⇒ where n = 3q is any integer)
a = 9q + 1 = 3(3q) + 1 = 3n + 1
(where n = 3q is any integer)
a = 9q + 2 = 3(3q) + 2 = 3n + 2
(where n = 3q is any integer)
Similarly a = 9q + 3 = 3(3q + 1) = 3n
(where n = 3q + 1 is any integer)
a = 9q + 4 = 3(3q + 1) + 1 = 3n + 1
(where n = 3q + 1 is any integer)
a = 9q + 5 = (9q + 3) + 2
= 3(3q + 1) + 2 = 3n + 2
(where n = 3q + 1 is any integer)
a = 9q + 6 = 3(3q + 2) = 3n
(where n = 3q + 2 is any integer)
a = 9q + 7 = 3(3q + 2) + 1 = 3n + 1
(where n = 3q + 2 is any integer)
a = 9q + 8 = 3(3q + 2) + 2 = 3n + 2
(where n = 3q + 2 is any integer)
So, all numbers. a = 9q, 9q + 1, 9q + 2, 9q + 3,
9q + 4, 9q + 5, 9q + 6, 9q + 7, 9q + 8 are expressed as
3n, 3n + 1, 3n + 2.
So, a = 3n, 3n + 1 or 3n + 2
Cube of all these numbers.
{tex}\style{font-family:Arial}{\begin{array}{l}a^3\;=\;(3n)^3\;=\;9(3n^3)\;=\;9m\;(where\;m\;=\;3n^3\;is\;any\;integer)\end{array}}{/tex}

{tex}\style{font-family:Arial}{\begin{array}{l}a^3=(3n+1)^3=(3n)^3+1^3+3(3n)(3n+1)\\=9(3n^3+3n^2+n)+1\\=9m+1\;(where\;m=3n^3+3n^2+n\;is\;any\;integer)\end{array}}{/tex}

{tex}\style{font-family:Arial}{a^3=(3n+2)^3=(3n)^3+(2)^3+3(3n)(2)(3n+2)\\=9(3n^3+6n^2+4n)+8=9m+8(where\;m\\=3n^3+6n^2+4n\;is\;any\;integer)}{/tex}

So, cube of any positive integer is of the form 9m, 9m + 1, 9m + 8.