To prove that $n^3 + 2n$ is divisible by 3 for all $n > 0$ and $n \in \mathbb{Z}^+$, we can approach this problem by examining $n$ modulo 3.

For any integer $n$, it will satisfy one of the following cases in modulo 3:

1. $n \equiv 0 \pmod{3}$
2. $n \equiv 1 \pmod{3}$
3. $n \equiv 2 \pmod{3}$

We will analyze each case to check if $n^3 + 2n \equiv 0 \pmod{3}$, ensuring it is divisible by 3 in all scenarios.

Case 1: $n \equiv 0 \pmod{3}$

If $n \equiv 0 \pmod{3}$, then: $n^3 \equiv 0^3 \equiv 0 \pmod{3}$ and $2n \equiv 2 \cdot 0 \equiv 0 \pmod{3}.$ Thus, $n^3 + 2n \equiv 0 + 0 \equiv 0 \pmod{3}.$ In this case, $n^3 + 2n$ is divisible by 3.

Case 2: $n \equiv 1 \pmod{3}$

If $n \equiv 1 \pmod{3}$, then: $n^3 \equiv 1^3 \equiv 1 \pmod{3}$ and $2n \equiv 2 \cdot 1 \equiv 2 \pmod{3}.$ Thus, $n^3 + 2n \equiv 1 + 2 \equiv 3 \equiv 0 \pmod{3}.$ In this case, ( n