Let $$n \geq 5$$ be an integer. If $$9^n - 8n - 1 = 64\alpha$$ and $$6^n - 5n - 1 = 25\beta$$, then $$\alpha - \beta$$ is equal to:

We are given: $$9^n - 8n - 1 = 64\alpha$$ and $$6^n - 5n - 1 = 25\beta$$, for $$n \geq 5$$. Expanding $$9^n = (1+8)^n$$ by the binomial theorem yields $$9^n = \sum_{k=0}^{n} \binom{n}{k} 8^k = 1 + 8n + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \ldots + 8^n,$$ so $$9^n - 8n - 1 = \binom{n}{2}8^2 + \binom{n}{3}8^3 + \ldots + \binom{n}{n}8^n = 64\Bigl[\binom{n}{2} + \binom{n}{3}\cdot 8 + \binom{n}{4}\cdot 8^2 + \ldots + \binom{n}{n}\cdot 8^{n-2}\Bigr],$$ and therefore $$\alpha = \sum_{k=2}^{n}\binom{n}{k}\cdot 8^{k-2}.$$ Similarly, expanding $$6^n = (1+5)^n$$ gives $$6^n - 5n - 1 = \binom{n}{2}5^2 + \binom{n}{3}5^3 + \ldots + \binom{n}{n}5^n = 25\Bigl[\binom{n}{2} + \binom{n}{3}\cdot 5 + \binom{n}{4}\cdot 5^2 + \ldots + \binom{n}{n}\cdot 5^{n-2}\Bigr],$$ so $$\beta = \sum_{k=2}^{n}\binom{n}{k}\cdot 5^{k-2}.$$ Computing $$\alpha - \beta$$ we have $$\alpha - \beta = \sum_{k=2}^{n}\binom{n}{k}(8^{k-2} - 5^{k-2}).$$ Since for $$k = 2$$ the term is zero, the sum effectively starts from $$k = 3$$: $$\alpha - \beta = \sum_{k=3}^{n}\binom{n}{k}(8^{k-2} - 5^{k-2}) = \binom{n}{3}(8^1 - 5^1) + \binom{n}{4}(8^2 - 5^2) + \ldots + \binom{n}{n}(8^{n-2} - 5^{n-2}).$$ This matches Option C: $$\binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \ldots + \binom{n}{n}(8^{n-2}-5^{n-2}).$$ Therefore, the answer is Option C.