TITLE: Show that $n^3+2n$ is divisible by 3 for all $n\ge 1$
QUESTION [5 upvotes]: i want to prove it with mathematical induction :
first i am tried with n=0
then it is divisible by zero then i move to next step change all n with K then i get this product :
$$P(K)=K^3+2K = 3m$$
Note: $3k$ because we multiply any no. with $3$ is divisible by $3$
now , next step i am increase $+1$ in $k$ so i get this step :
$$(K+1)^3 + 2(K+1) = 3m$$
so now next step i am not able to solve please help.

REPLY [3 votes]: $\begin{eqnarray}{\bf Hint}\ \ \ {\rm mod}\ 3\!:\,\ \color{#0a0}{n^3}\! &\equiv& \color{#c00}n\,\ (\equiv\, -2n)  &&{\rm i.e.}\ \ \ P(n)\\ \Rightarrow\  (n\!+\!1)^3\! &=& \color{#0a0}{n^3}\! + \color{lightgrey}{3n^2\!+3n}\!+\!1\\ &\equiv& \color{#c00}n + 1 && {\rm i.e.}\ \ \ P(n\!+\!1)\end{eqnarray}$