Proving $n^3 + 3n^2 +2n$ is divisible by $6$ The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.
So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not. 
 A: $n^3+3n^2+2n=(n+1)(n+2)n$. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both $2$ and $3$ and hence is divisible by the least common multiple of $2$ and $3$, which turns out to be $6$.
A: The product of three consecutive integers is divisible by $2$ since one of the factors need to be even. It is also divisible by $3$. (why?) The  the product of three consecutive integers is then divisible by the least common multiple of those two numbers $2\times 3 =6$.
Alternatively, note that $$n(n+1)(n+2) = 6 {n+2 \choose 3}$$
A: $$ n^3+3n^2+2n = 6\binom{n+2}{3} \in 6\mathbb{Z}.$$
A: Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.
Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$
$$\begin{array}{c | c c c}
n & n+1 & n+2 & n(n+1)(n+2)\\
\hline
0 & 1 & 2 & 0\\
1 & 2 & 0 & 0\\
2 & 0 & 1 & 0
\end{array}$$
So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$
One could also test directly the congruences modulo $6$
$$\begin{array}{c|c c c c}
n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\
\hline
0 & 0 & 0 & 0 & 0\\
1 & 1 & 3 & 2 & 0\\
2 & 2 & 0 & 4 & 0\\
3 & 3 & 3 & 0 & 0\\
4 & 4 & 0 & 2 & 0\\
5 & 5 & 3 & 4 & 0
\end{array}$$
And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$
A: Using only modular arithmetic, without factoring, you can see that with $p(n)=n^{3}+3 n^{2}+2 n$ we have if $n\cong0(mod 2)$ then $p(n)\cong0+0+0=0 (mod 2)$ if $n\cong1(mod 2)$ then $p(n)\cong1+3+2\cong0(mod 2)$ So $2|p(n)$. Similarly if $n\cong1(mod 3)$ then $p(n)\cong1+3+2\cong0(mod 3)$, if $n\cong2(mod 3)$ then $p(n)\cong8+12+4\cong0(mod 3)$ so $3|p(n)$. Since both 2 and 3 divide $p(n)$ then $p(n)$ is a multiple of 6.
