⚡ AZUCATION SHORTCUT:
Check Domain first: $x-3 > 0$ and $x-3 \neq 1 \implies x > 3, x \neq 4$.
Simplify: $(x-3)(x+3) = (x+1)(x-3)^2$. Divide by $(x-3)$: $x+3 = (x+1)(x-3)$.
Solve $x^2-3x-6=0 \implies x = \frac{3 \pm \sqrt{33}}{2}$. Only $\frac{3+\sqrt{33}}{2}$ satisfies $x > 3$.

Step 1: Determine the Domain
For the logarithm $\log_{base}(arg)$ to be defined:
1. Base must be positive and not 1: $x-3 > 0 \implies x > 3$ and $x-3 \neq 1 \implies x \neq 4$.
2. Arguments must be positive: $x^2-9 > 0 \implies x > 3$ or $x < -3$. Also, $x+1 > 0 \implies x > -1$.
Intersection: $x > 3$ and $x \neq 4$.

Step 2: Solve the Equation
Using $\log_a b = c \iff b = a^c$ or logarithmic properties:
$\log_{x-3}(x^2 - 9) = \log_{x-3}(x+1) + \log_{x-3}(x-3)^2$
$\log_{x-3}(x^2 - 9) = \log_{x-3}[(x+1)(x-3)^2]$
$x^2 - 9 = (x+1)(x-3)^2$
$(x-3)(x+3) = (x+1)(x-3)^2$

Since $x \neq 3$ from the domain, divide by $(x-3)$:
$x+3 = (x+1)(x-3)$
$x+3 = x^2 - 2x - 3$
$x^2 - 3x - 6 = 0$

Step 3: Validate Roots
Using quadratic formula: $x = \frac{3 \pm \sqrt{9 - 4(1)(-6)}}{2} = \frac{3 \pm \sqrt{33}}{2}$.
$x_1 = \frac{3 + \sqrt{33}}{2} \approx 4.37$ (Valid as it is $>3$ and $\neq 4$).
$x_2 = \frac{3 - \sqrt{33}}{2} \approx -1.37$ (Invalid as it is $<3$).
Only one valid value exists, so the sum is $\mathbf{\frac{3 + \sqrt{33}}{2}}$.

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