Let the coefficient of $$x^r$$ in the expansion of $$(x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \ldots + (x+2)^{n-1}$$ be $$\alpha_r$$. If $$\sum_{r=0}^{n}\alpha_r = \beta^n - \gamma^n$$, $$\beta, \gamma \in \mathbb{N}$$, then the value of $$\beta^2 + \gamma^2$$ equals

 $$a = (x+3)^{n-1}$$

$$ r = \frac{x+2}{x+3} $$

$$ S = a \cdot \frac{1 - r^n}{1 - r} $$

$$ S = (x+3)^{n-1} \cdot \frac{1 - \left(\frac{x+2}{x+3}\right)^n}{1 - \frac{x+2}{x+3}} $$

$$ 1 - \frac{x+2}{x+3} = \frac{(x+3) - (x+2)}{x+3} = \frac{1}{x+3} $$

$$ 1 - \left(\frac{x+2}{x+3}\right)^n = \frac{(x+3)^n - (x+2)^n}{(x+3)^n} $$

$$ S = (x+3)^{n-1} \cdot \frac{(x+3)^n - (x+2)^n}{(x+3)^n} \cdot (x+3) = (x+3)^n - (x+2)^n $$

To find the sum of all coefficients of a polynomial $$f(x)$$, we substitute $$x = 1$$

$$ \sum_{r=0}^{n} \alpha_r = S(1) = (1+3)^n - (1+2)^n = 4^n - 3^n $$

This is because when we substitute $$x = 1$$, every term $$\alpha_r x^r$$ becomes just $$\alpha_r$$, so their sum equals the polynomial evaluated at $$x = 1$$.

Finally, comparing $$4^n - 3^n = \beta^n - \gamma^n$$ with $$\beta, \gamma \in \mathbb{N}$$, we identify

$$\beta = 4$$ and $$\gamma = 3$$.

$$ \beta^2 + \gamma^2 = 16 + 9 = 25 $$