To prove that for all $k \in \mathbb{N}$, $k^3 + 2k$ is divisible by 3, we will use mathematical induction.

Step 1: Base Case

We need to verify the statement for $k = 1$.

For $k = 1$:

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

Since $3$ is divisible by 3, the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement holds for some arbitrary $k = n$, i.e.,

$n^3 + 2n \text{ is divisible by 3.}$

This means that there exists some integer $m$ such that:

$n^3 + 2n = 3m \quad \text{(1)}$

Step 3: Inductive Step

Now, we must prove that the statement holds for $k = n + 1$, i.e., we want to show that:

$(n+1)^3 + 2(n+1) \text{ is divisible by 3.}$

First, expand $(n+1)^3 + 2(n+1)$:

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

Simplifying:

$(n+1)^3 + 2(n+1) = n^3 + 3n^2 + 3n + 1 + 2n + 2 = n^3 + 3n^2 + 5n + 3$

Now, we can factor this expression:

$n^3 + 3n^2 + 5n + 3 = (n^3 + 2n) + 3n^2 + 5n + 3$

From the inductive hypothesis, we know that $n^3 + 2n = 3m$ for some integer $m$. Substituting this in:

$n^3 + 3n^2 + 5n + 3 = 3m + 3n^2 + 5n + 3$

Now, factor out the 3 from the terms that are clearly divisible by 3:

$3m + 3n^2 + 5n + 3 = 3(m + n^2 + \frac{5n}{3} + 1)$

Since we know all the terms are divisible by 3. We conclude the induction step