The problem asks us to use a direct proof to show that the cube of every even integer is even.

1. Problem Description

The problem asks us to use a direct proof to show that the cube of every even integer is even.

2. Solution Steps

Let $n$ be an arbitrary even integer.

By definition, an even integer can be expressed as $2k$ for some integer $k$.

So, $n = 2k$.

Now, we want to find the cube of $n$, which is $n^3$.

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

$n^3 = 2^3 * k^3$

$n^3 = 8k^3$

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

Let $m = 4k^3$. Since $k$ is an integer, $4k^3$ is also an integer. Thus, $m$ is an integer.

Then, $n^3 = 2m$, where $m$ is an integer.

Since $n^3$ can be written as $2$ times an integer, $n^3$ is an even integer.

3. Final Answer

Therefore, the cube of every even integer is even.