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

Let a be any positive integer and b = 3

a = 3q + r, where q \(\ne\) 0 and 0 < r < 3

a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented as these three forms.

We have three cases.

Case 1 : When a = 3q,
\(a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m \)

Where m is an integer such that \(m = 3q^3\)
Case 2 :When a = 3q + 1
\(\Rightarrow \) \(a^3 = (3q + 1)^3\)
\(\Rightarrow \) \(a^3 = 27q^3 + 27q^2 + 9q + 1\)
=> \(a^3 = 9( 3q^3 + 3q^2 + q) + 1\)
\(\Rightarrow \) \(a^3 = 9m + 1\)

When m is an integer such that \(m = (3q^3 + 3q^2 + q)\)

Case 3 :When a = 3q + 2,
\(\Rightarrow \) \(a^3 = (3q + 2)^3\)
\(\Rightarrow \) \(a^3 = 27q^3 + 54q^2 + 36q + 8\)
\(\Rightarrow \) \(a^3 = 9( 3q^3 + 6q^2 + 4q) + 8\)
\(\Rightarrow \) \(a^3 = 9m + 8\)

When m is an integer such that \(m = (3q^3 + 6q^2 + 4q)\)

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