Partial Fraction Decomposition with Exponents

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I want to break this fraction into two parts: $$\frac{5^n3^n}{(5^n-3^n)(5^{n+1}-3^{n+1})}$$ when I try to equate this to: $$\frac{A}{(5^n-3^n)} - \frac{B}{(5^{n+1}-3^{n+1})} $$ I get the value of A by substituting the value of n by 0: $$A(5^{0+1}-3^{0+1}) = 5^03^0 $$ $$ A= \frac{1}{2} $$ Similarly the value of B is -0.5

By another method when I try to find the value of A and B $$5^n(5A + B) - 3^n(3A+B) = 5^n3^n$$ and then solving for 5A+B = 3^n and 3A + B = 0
I get the value of: $$A = \frac{3^n}{2}$$ $$B = -\frac{3^{n+1}}{2}$$

I can't understand why both methods give different values and why the first one is wrong?

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Partial fractions is for rational functions of a variable $x$, and what you have is not a rational function of $n$. The methods that work for rational functions should not be expected to work here.

If you want to write $$ \frac{5^n 3^n}{(5^n- 3^n)(5^{n+1}-3^{n+1})} = \frac{A}{5^n - 3^n} - \frac{B}{5^{n+1}-3^{n+1}}$$ (presumably when the none of the denominators are $0$), that is equivalent to $$ 5^n 3^n = (5^{n+1}-3^{n+1}) A - (5^n-3^n) B $$ Now there are certainly no solutions where $A$ and $B$ are constants. A solution where $A$ and $B$ are multiples of $3^n$ is $A= 3^n/2$, $B = 3^{n+1}/2$. A solution where $A$ and $B$ are multiples of $5^n$ is $A = 5^n/2$, $B = 5^{n+1}/2$. Or you could take a combination of these solutions:

$$ A = \frac{5^n}{2} t + \frac{3^n}{2} (1-t),\ B = \frac{5^{n+1}}{2} t + \frac{3^{n+1}}{2} (1-t) $$

where $t$ is arbitrary.

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$n=0$ is not the case you want to work with. $5^0=3^0=1$. Therefore

$\displaystyle \frac{A}{(5^0 - 3^0)} = $???

Note that if you solve for any particular $n$ (other than zero) you will get results that are correct for that particular $n$, but not correct for the general problem.