prove for all $n\geq 0$ that $3 \mid n^3+6n^2+11n+6$ I'm having some trouble with this question and can't really get how to prove this..
I have to prove $n^3+6n^2+11n+6$ is divisible by $3$ for all $n \geq 0$. 
I have tried doing $\dfrac{m}{3}=n$ and then did $m=3n$
then I said $3n=n^3+6n^2+11n+6$ but now I am stuck.
 A: What you can do is the following: Every integer in the universe is either of the form $3m, 3m+1 $ or $3m+2$. Then do a casebash:
If $n=3m$, then clearly $3|(3m)^3+6(3m)^2+11(3m) + 6$.
If $n = 3m+ 1$, then putting this into your expression above we get 
$$\begin{eqnarray} (3m+1)^3 + 6(3m+1)^2 + 11(3m+1) + 6 &=& 3(\text{stuff}) + 24 \\
&=& \text{a multiple of 3}. \end{eqnarray}$$
You can do a similar computation for $n= 3m+2$ as well so you are done.
A: Here is a solution using induction:
Let $f(x)=x^3+6x^2+11x+6$
Since we want to see if it is divisible by 3 let us assume that $f(x)=3m$.
For the case where $x=0$, $f(0)=6$ which is divisible by 3. 
Now that we have proved for one case let us prove for the case of $f(x+1)$
$$f(x+1)=(x+1)^3+6(x+1)^2+11(x+1)+6$$
$$= x^3+3x^2+3x+1+6x^2+12x+6+11x+11+6$$
$$=(x^3+6x^2+11x+6)+3x^2+15x+18$$
And since $x^3+6x^2+11x+6=3m$
$$f(x+1)=3m+3x^2+15x+18=3(m+x^2+5x+6)$$
Which is divisible by 3.
A: If you know what "mod 3" means then argue as follows:
$$n^3 + 6n^2 + 11n + 6 \equiv n^3 - n = (n-1)n(n+1) \equiv 0 \pmod 3 .$$
If you don't, then write this as:
$$ n^3 - n + 12n + 6n^2 + 6 = n(n+1)(n-1) + 3(2n^2 + 4n + 2), $$
and you're left with showing that both terms are divisible by $3$.
Now $n(n+1)(n-1)$ is always a multiple of $3$, because if a number is not a multiple of 3, then either its predecessor or its successor must be.
A: We have
$$n^3 + 6n + 11n + 6 = (n+1)(n+2)(n+3).$$
A run of three consecutive integers has an integer in it divisible by 2, so 
$$2|n^3 + 6n + 11n + 6.$$   A run of three consecutive integers has an integer in it divisible by 3, so 
$$3|n^3 + 6n + 11n + 6.$$ 
Since 2 and 3 are relatively prime, the  desired conclusion follows right away.
A: We have
$$
\begin{align}
n^3+6n^2+11n+6
&=6\binom{n}{3}+18\binom{n}{2}+18\binom{n}{1}+6\binom{n}{0}\\
&=6\left(\binom{n}{3}+3\binom{n}{2}+3\binom{n}{1}+\binom{n}{0}\right)
\end{align}
$$
so $6\mid(n^3+6n^2+11n+6)$ for all $n\in\mathbb{Z}$.
Of course, since $3\mid 6$, we have $3\mid(n^3+6n^2+11n+6)$, as requested.
A: There's an algorithm for finding rational roots of a polynomial with integer coefficients, and if you use that to factor this polynomial you get
$$
n^3 + 6n^2 + 11n + 6 = (n+1)(n+2)(n+3).
$$
For any $n$, one of those three factors is a multiple of $3$.
A: Prove it for $n=0,1,2$, which is a simple calculation. Every number is of the form $n=3x+r$ where $r=0,1,2$. Now expand $n^3+6n^2+11n+6$ in terms of $x$ and $r$. You'll get $r^3+6r^2+11r+6$ plus a multiple of $3$.
A: $n^3+6n^2+11n+6=n^3+n^2+5n^2+5n+6n+6=n^2(n+1)+5n(n+1)+6(n+1)=$
$=(n+1)(n^2+5n+6)=(n+1)(n^2+3n+2n+6)=(n+1)(n+2)(n+3)$
Now , since this last expression represents a product of three consecutive integers it has to be divisible by $3$ .
A: A simple fact to remember is that the "modulo k" operator can be copied "inside" addition, subtraction and multiplication without altering the result... to say it more formally
$$(a + b)_{mod\ k} = (a_{mod\ k} + b_{mod\ k})_{mod\ k}$$
$$(a - b)_{mod\ k} = (a_{mod\ k} - b_{mod\ k})_{mod\ k}$$
$$(ab)_{mod\ k} = ((a_{mod\ k})(b_{mod\ k}))_{mod\ k}$$
$$(a^n)_{mod\ k} = ((a_{mod\ k})^n)_{mod\ k}$$
Your problem is to prove that $(n^3 + 6n^2 + 11n + 6)_{mod\ 3} = 0$. Just expanding the problem statement using the above equations you get:
$$((n_{mod\ 3})^3 + (6_{mod\ 3})(n_{mod\ 3})^2 + (11_{mod\ 3})(n_{mod\ 3}) + 6_{mod\ 3})_{mod\ 3} = 0$$
Observing that $6_{mod\ 3}=0$ and $11_{mod\ 3}=2$ the expression can be simplified to
$$ ((n_{mod\ 3})^3 + 2(n_{mod\ 3}))_{mod\ 3} = 0 $$
Given that $n_{mod\ 3}$ is either 0, 1 or 2 you only need to test these three cases and this is trivial as it amounts to checking that $(0+0)$, $(1+2)$ and $(8+4)$ are all divisible by 3.
Note of course that it was also quite obvious without any computation that adding or subtracting $6$ or $6n^2$ (both multiples of 3) was not going to change the fact that an expression is divisible by 3 or not.
