How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3. 
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
 A: You've already solved the case $n=1$, so I'll not repeat that there.
Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain
$$
\begin{split}
(n+1)((n+1)^2+8) & = (n+1)(n^2+2n+1+8)\\
& =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\
& = n(n^2+8)+3n^2+3n+9
\end{split}
$$
In the latter, all the terms are divisible by $3$, hence it follows that $(n+1)((n+1)^2+8)$ is also divisible by $3$. This finishes the induction proof, so we may conclude that $n(n^2+8)$ is divisible by $3$ for all $n\geq 1$.
A: Without induction: 
Since $8=-1\pmod{3}$, $n(n^2+8)=n(n^2-1)=(n-1)n(n+1)\pmod{3}$. Since $n-1$, $n$ and $n+1$ are three consecutive integers, at least (and in fact, exactly) one of them is a multiple of $3$, hence their product is a multiple of $3$.
A: Hint $\rm\  n\:\!(n^2\!-\!1+9) = (n\!-\!1)\:\!n\:\!(n\!+\!1) + 9\:\!n\:$ so it suffices to show $3$ divides one of $\rm\:\!n\!-\!1, n, n\!+\!1. $
The base case $\rm\:n=1\:$ is true since $3$ divides $\rm\:n\!-\!1 = 0.\:$ For the induction step notice that
$\quad 3$ divides one of $\rm\: n\!-\!1, n, n\!+\!1\:\Rightarrow\: 3$ divides one of $\rm \:n,n\!+\!1,n\!+\!2\:\ $ by $\rm\:\ n\!+\!2\:\! =\:\! n\!-\!1 + 3$
For more general methods see my many posts on telescopy.
A: If $n\equiv 0\pmod 3$ Ok. If $n\equiv 1\pmod 3$, we have
\begin{equation}
n^{2} + 8 \equiv 1^{2} + 2\equiv 0\pmod 3.
\end{equation}
If $n \equiv 2\pmod 3$ we have
\begin{equation}
n^{2} + 8 \equiv 2^{2} + 2\equiv 6 \equiv 0\pmod 3.
\end{equation}
A: You start by supposing that $n(n^2+8)$ is a multiple of 3, and you need to show that $(n+1)((n+1)^2 + 8)$ is also a multiple of 3. I would start by simplifying the latter expression.  
Then manipulate the expression until it is a sum of things that are multiples of 3. This includes expressions like $3a$, which is always a multiple of 3 for any $a$, and  $n(n^2+8)$, by the induction hypothesis.
A: Using induction it is obvious that the statement is true for $n=1$. 
Now suppose that it is true for $n=k$,
then we have $k(k^2+8)=3m$, where m is an integer.
Considering the case where $n=k+1$, we have;
$$(k+1)[(k+1)^2+8]=k(k^2+2k+9)+k^2+2k+9$$
$$=k(k^2+8+2k+1)+k^2+2k+9=k(k^2+8)+k(2k+1)+k^2+2k+9$$
$$=k(k^2+8)+3k^2+3k+9=3m+3k^2+3k+9=3(m+k^2+k+3)$$
A: If $(n+1)((n+1)^2+8)=(n+1)(n^2+2n+9)$ then if $(n+1)= 0\  mod\ 3$, we're done. If not, than $(n+1)=1\mod 3$ or $(n+1)= 2\mod 3$. So if $n+1=1\mod 3 $ then $n=0\mod 3$ so $3|(n^2+2n+9)$, and if $n+1=2\mod 3$ then $n^2=n=1\mod 3$ now let $n^2=3k+1$ and $n=3l+1$ so we have $(n^2+2n+9)=(3k+1+6l+2+9)=3(k+2l+1+3)$. 
A: Since you have proven that this formula is available for $n=1$ we suppose that formula is available for all natural numbers n=k
$$k(k^2+8)$$ then according to axiomme of mathematical induction we need to prove that formula is valid for $n=k+1$ or 
$$(k +1)((k+1)^2+8)$$ is multiple of 3
now we have 
$$(k+1)((k+1)^2+8)=(k+1)(k^2+2k+1+8)=k^3+3k^2+11k+9=(k^3+8k)+(3k^2+3k+9)=k(k^2+8)+3(k^2+k+3)$$
first part of expression is factor of 3 by assumption an second part evidently is factor of 3
A: Quote: How can I show that $3n^2 + 3n + 9$ is a multiple of $3$? End of quote
I'm surprised at the complexity of some of the answers.
$$
3n^2 + 3n + 9 = 3(n^2 + n + 3) = (3\cdot\text{something}).
$$
A "multiple of $3$" is anything that is $3$ times something (where "something" means of course an integer).
