Show that $n^3+2n$ is divisible by 3 for all $n\ge 1$ 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.
 A: For the induction step:
$(n+1)^3+2(n+1)=n^3+2n+3n+3n^2+3=n^3+2n+3(n+n^2+1)$ 
$n^3+2n$ is divisible by 3 (by assumption) and the last addend is obviusly divisible by 3
Remark: If you want to show a statement for all $n\geq k$ then you prove the statement in the first step for k, not for $0$. (In our case we wont get any problems)
A: $$\left((n+1)^3+2(n+1)\right)-\left(n^3+2n\right) = 3(n^2+n+1)$$
is always a multiple of three, hence if $3\mid (n^3+2n)$, then $3\mid ((n+1)^3+2(n+1)).$
A: An alternate easier solution: Since $n^3+2n = n(n^2+2) \equiv n(n^2-1) = (n-1)n(n+1) \pmod 3$, for any $n$, either $n$ or $n-1$ or $n+1$ is zero modulo $3$ and so their product is divisible by $3$. 
A: $\begin{eqnarray}{\bf Hint}\ \ \ {\rm mod}\ 3\!:\,\ \color{#0a0}{n^3}\! &\equiv& \color{#c00}n\,\ (\equiv\, -2n)  &&{\rm i.e.}\ \ \ P(n)\\ \Rightarrow\  (n\!+\!1)^3\! &=& \color{#0a0}{n^3}\! + \color{lightgrey}{3n^2\!+3n}\!+\!1\\ &\equiv& \color{#c00}n + 1 && {\rm i.e.}\ \ \ P(n\!+\!1)\end{eqnarray}$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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Lets $\ds{n \equiv 3p + \delta}$ for an integer $\ds{p}$ where $\ds{\delta \in \braces{0,1,2}}$. Then,

\begin{align}
n^{3} + 2n&
=\pars{27p^{3} + 27p^{2}\delta + 9p\delta^{2} + \delta^{3}} + \pars{6p + 2\delta}
\\[3mm]&=\color{#c00000}{\Large 3}
\pars{9p^{3} + 9p^{2}\delta + 3p\delta^{2} + 2p}
+
\underbrace{\pars{\delta^{3} + 2\delta}}
_{\ds{\in\ \color{#c00000}{\Large\braces{0,3,12}}}}
\end{align}

A: Alternatively: $n^3+2n\equiv n^3-n\equiv 0 \pmod 3$, using Fermat's little theorem.
A: I know you want to prove by induction, but anothter proof would be to observe that every integer is congruent to either 0, 1, or 2 (mod 3). Then we want to see if for every integer n:
$n^3 + 2n \equiv 0\pmod{3}$
Well, take all integers n such that $n \equiv 0\pmod{3}$. Then $n^3 \equiv 0\pmod{3}$ and $2n \equiv 0\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 1\pmod{3}$. Then $n^3 \equiv 1\pmod{3}$ and $2n \equiv 2\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 2\pmod{3}$. Then $n^3 \equiv 2\pmod{3}$ and $2n \equiv 4\equiv1\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
A: $$it-is--a-proof-without-induction\\\\M=n^3+2n =\\n^3+2n-3n +(3n)\\=n^-n +(3n) \\=n(n^2-1)=(3n)\\=n(n-1)(n+1) +3n \\=3k +3n\\=3(k+n)=3q
$$
