Answer
Let $P(n)$ denote the proposition: "3 divides $n^3 +2n$ whenever $n$ is a positive integer."
$P(1)$ is true, since $(1)^3 +2(1) = 3$.
Assume that $P(k)$ is true for an arbitrary positive integer $k$. Now we must show that this implies that $P(k+1)$ is also true.
\begin{align*}
(k+1)^3 + 2(k+1) &= k^3 + 3k^2 + 3k + 1 +2k +2 \\
&= k^3 + 2k + 3k^2 + 3k +3 \\
&= [k^3 +2k] + [3k^2] + [3k] + [3]
\end{align*}
Each of the bracketed terms is divisible by $3$, so $P(k+1)$ is also true. Therefore by method of mathematical induction $P(n)$ is true - 3 divides $n^3 +2n$ whenever $n$ is a positive integer.
Work Step by Step
First, we must give the proposition. We let $P(n)$ denote the proposition: "$3$ divides $n^3 +2n$ whenever $n$ is a positive integer."
Next, we must prove our base case $P(1)$ is true by showing that the proposition is true when $n=1$. When $n=1$, $n^3 +2n = (1)^3 +2(1) =3$. $3$ is divisible by $3$, so therefore $P(1)$ is true.
Finally, we must carry out the inductive step, Assume that $P(k)$ is true; that is, we assume that $k^3 +2k$ is divisible by $3$ for an arbitrary positive integer $k$. Now we must show that this implies that $P(k+1)$ is also true.
\begin{align*}
(k+1)^3 + 2(k+1) &= (k+1)(k^2 +2k +1) +2k +2 \\
&= k^3 + 3k^2 + 3k + 1 +2k +2 \\
&= k^3 +3k^2 +2k + k +1 +2k +2 \\
&= k^3 + 2k + 3k^2 + 3k +3 \\
&= [k^3 +2k] + [3k^2] + [3k] + [3]
\end{align*}
The first bracketed term is divisible by $3$, via our inductive hypothesis; the rest of the bracketed terms are trivially divisible by $3$. This means that the entire expression must also be divisible by $3$, and that $P(k+1)$ is true. Therefore by method of mathematical induction $P(n)$ is true - 3 divides $n^3 +2n$ whenever $n$ is a positive integer.
