Prove by induction: $\forall n\in\mathbb{Z}_{\geq1}:3\ |\ (6n^2-12n+3)$ I'm not sure how to start this induction problem.

I  was told that we start doing induction by using a base case $n=1$.  Then we set $n=k$ to prove $n=k+1$.  But how do I prove that $6(k+1)^2-12(k+1)+3$ is divisible by $3$ if $6k^2-12k+3$ is?

I'm confused, can someone give me some help?  Thanks.
Hint:  If you have to use induction, you have the correct idea.  You assume $3\mid6k^2-12k+3$ and want to prove $3\mid6(k+1)^2-12(k+1)k+3$.  You make use of your assumption by finding it in the new expression.  So expand out the $(k+1)$'s and group the terms together to use the assumption.  Now you look at what is left and show that $3$ divides it, too.
You can also proceed as follows: 

Prove this for $n=1$.

Assume $3|(6k^2-12k+3)$.

Then $6(k+1)^2-12(k+1)k+3 = 3(2(k+1)^2-4(k+1)+1)$ which is divisible by $3$.

And you're done.

The idea of induction is that you assume $P(n)$ is true for some $k$ and then prove it for $k+1$. If you prove it for $k+1$ without using the induction hypothesis, it is still a valid induction argument since you first assumed $P(n)$ is true for some $k$ and then proved it for $k+1$.