$S_n = \frac{12}{(4-3)(4^2-3^2)} + \frac{12^2}{(4^2-3^2)(4^3-3^3)} + ... + \frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$ 
So I know that in order to find the series we need to change the form because we can cancel the terms  $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} = \frac{A}{4^n-3^n} + \frac{B}{4^{n+1}-3^{n+1}}$$
But my question is how do we get $A = \frac{4^n+3^n}{2}$ and $B = \frac{4^{n+1}+3^{n+1}}{2}$
When I find the common denominator and add them together I got $12^n = A(4^{n+1}-3^{n+1}) + B(4^n-3^n)$. I tried simplifying but it can't seem to work.
Can anyone show me how to find A and B? Thank you in advance.

$$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$$
$4^n=a$, $3^n=b$, $12^n=ab$
$$\frac{ab}{(a-b)(4a-3b)}$$
Partitional fractioning in two variables is not unique. Make polynomial partitional fractioning considering $a$ as independent variable
$$\frac{ab}{(a-b)(4a-3b)}=\frac{A}{a-b}+\frac{B}{4a-3b}$$
$$ab=4aA-3bA+Ba-Bb=(4A+B)a-(3A+B)b$$
$$4A+B=b, 3A+B=0 \Rightarrow A=b, B=-3b$$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{b}{a-b}-\frac{3b}{4a-3b}$$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{b}{a-b}+\frac{1}{2}-\left(\frac{3b}{4a-3b}+\frac{1}{2}\right)$$
Combining $\frac{b}{a-b}+\frac{1}{2}$ and $\frac{3b}{4a-3b}+\frac{1}{2}$
$$\frac{ab}{(a-b)(4a-3b)}=\frac{a+b}{2(a-b)}-\frac{4a+3b}{2(4a-3b)}$$
