Find the value of $abc$. 
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.

In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?

Let $M=396396$. Because $599\times 499<M$, we infer that $a>5$. Because $11|M$, we infer that $a+c-b=11$. Finally, $3|M$ and so $3|(a+b+c)$. So let's consider the possibilities for $a+b+c$:
$$
a+b+c=2b+11\in\{12,15,18,21,24,27\}
$$
which yields $b\in\{2,5,8\}$ which corresponds to $a+c\in\{13,16,19\}$. $19$ is too high because $c<a$. If $b=5$, then it must be that $a=9$ and $b=7$. You can verify that this doesn't work. Thus
$$
b=2\implies a+c=13\implies (a,c)\in\{(7,6),(8,5),(9,4)\}.
$$
Only $\boxed{abc=924}$ works and that's our answer.
Edit: I left out the case $a+c=b$. Let's look at it
$$
a+b+c=2b\in\{0,6,12,18\}\implies b\in\{0,3,6,9\}\implies a+c\in\{0,3,6,9\}.
$$
$0$ and $3$ are too small, and so $(a,c)\in\{(6,0),(6,3),(7,2),(8,1)\}$. Eliminate these by inspection.
