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?

Hint: 
Let $[xyz]$ denote a thre-digit number. Then you want $a,b,c$ such that $[abc]\cdot[cba]=396396$, with $a>c$.
Write $[abc]=100\cdot a+100\cdot b+1\cdot c$ and $[cba]=100\cdot c+10\cdot b+1\cdot a$, then
$$[abc]\cdot[cba]=(100\cdot a+100\cdot b+1\cdot c)\cdot(100\cdot c+10\cdot b+1\cdot a)$$
$$=10\ 000\cdot ac+1000\cdot(ab+bc)+100\cdot(a^2+b^2+c^2)+10\cdot(ab+bc)+ac$$
and this should equal $396396$. Can you continue from here?

Edit:
Listing these terms as follow:


*

*$10\ 000$: $ac$

*$1000$:  $\ \ \ ab+bc$   

*$100$: $\ \ \ \ \ a^2+b^2+c^2$

*$10$: $\ \ \ \ \ \ ab+bc$ 

*$1$: $\ \ \ \ \ \ \ \ ac$


You can see some patterns, e.g. the terms $ac$ and $ab+bc$ occur twice. Now compare this with the structure of $396396$ and consider the prime factorization of $396396$.
