Prove by strong induction that $3^n$ divides $a_n$ for all integers $n \ge 1$ 
Let $a_1 = 3, a_2 = 18$, and $a_n = 6a_{n-1} − 9a_{n-2}$ for each integer $n \ge 3$. Prove by strong induction that $3^n$ divides $a_n$ for all integers $n \ge 1$

I've done the base step and ih however I am stuck on the Inductive Step. I'm thinking it's something like $a_{k+1} = 6a_k -  9a_{k-1}$ but I don't know how to follow that.
Thanks in advance for the help with the Inductive Step.

Use strong induction: assume $a_{n - 2} = 3^{n - 2}\lambda_{n - 2}$ and $a_{n - 1} = 3^{n - 1}\lambda_{n - 1}$.  We then have:
\begin{align}
a_n =&\ 2\cdot 3\cdot 3^{n - 1}\lambda_{n - 1} - 3^2\cdot 3^{n - 2}\lambda_{n - 2}\\
=&\ 2\lambda_{n - 1}3^n - \lambda_{n - 2}3^n \\
=&\ \left(2\lambda_{n - 1} - \lambda_{n - 2}\right)3^n
\end{align}
$2\lambda_{n - 1} - \lambda_{n - 2}$ is surely an integer (integers are closed under addition and multiplication) therefore $a_n$ is a multiple of $3^n$.  The only thing left is to show that $a_1$ and $a_2$ satisfy this condition.
