The problem states that if $n$ is an odd integer, then $n^3$ is odd. We need to prove this statement.

1. Problem Description

The problem states that if $n$ is an odd integer, then $n^3$ is odd. We need to prove this statement.

2. Solution Steps

If $n$ is an odd integer, then it can be represented as:

$n = 2k + 1$, where $k$ is an integer.

Now we want to find $n^3$:

$n^3 = (2k + 1)^3$

$n^3 = (2k + 1)(2k + 1)(2k + 1)$

$n^3 = (4k^2 + 4k + 1)(2k + 1)$

$n^3 = 8k^3 + 4k^2 + 8k^2 + 4k + 2k + 1$

$n^3 = 8k^3 + 12k^2 + 6k + 1$

We can rewrite this as:

$n^3 = 2(4k^3 + 6k^2 + 3k) + 1$

Let $m = 4k^3 + 6k^2 + 3k$, where $m$ is an integer.

Then $n^3 = 2m + 1$

Since $n^3$ can be written in the form $2m + 1$, where $m$ is an integer, $n^3$ is an odd integer.

3. Final Answer

If $n$ is an odd integer, then $n^3$ is odd.