Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $$\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}}\right)^n$$, in the increasing powers of $$\frac{1}{\sqrt[4]{3}}$$ be $$\sqrt[4]{6} : 1$$. If the sixth term from the beginning is $$\frac{\alpha}{\sqrt[4]{3}}$$, then $$\alpha$$ is equal to _______

We have the binomial expansion of $$\left(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}}\right)^n$$ in increasing powers of $$\frac{1}{\sqrt[4]{3}}$$. The general term is $$T_{r+1} = \binom{n}{r}(\sqrt[4]{2})^{n-r}\left(\frac{1}{\sqrt[4]{3}}\right)^r = \binom{n}{r}\cdot 2^{(n-r)/4}\cdot 3^{-r/4}$$.

The fifth term from the beginning is $$T_5$$ (with $$r = 4$$): $$T_5 = \binom{n}{4}\cdot 2^{(n-4)/4}\cdot 3^{-1}$$.

The fifth term from the end is the $$(n+2-5)$$th = $$(n-3)$$th term from the beginning, which is $$T_{n-3}$$ (with $$r = n-4$$): $$T_{n-3} = \binom{n}{n-4}\cdot 2^{4/4}\cdot 3^{-(n-4)/4} = \binom{n}{4}\cdot 2\cdot 3^{-(n-4)/4}$$.

The ratio $$\frac{T_5}{T_{n-3}} = \frac{2^{(n-4)/4}\cdot 3^{-1}}{2\cdot 3^{-(n-4)/4}} = \frac{2^{(n-4)/4}}{2}\cdot\frac{3^{(n-4)/4}}{3} = \frac{2^{(n-4)/4}\cdot 3^{(n-4)/4}}{6} = \frac{6^{(n-4)/4}}{6}$$.

We are told this ratio equals $$\sqrt[4]{6} : 1 = 6^{1/4}$$. So $$\frac{6^{(n-4)/4}}{6} = 6^{1/4}$$, giving $$6^{(n-4)/4} = 6^{1+1/4} = 6^{5/4}$$.

Therefore $$\frac{n-4}{4} = \frac{5}{4}$$, so $$n - 4 = 5$$, hence $$n = 9$$.

Now we find the sixth term from the beginning, $$T_6$$ (with $$r = 5$$):

$$T_6 = \binom{9}{5}(\sqrt[4]{2})^{4}\left(\frac{1}{\sqrt[4]{3}}\right)^5 = \binom{9}{5}\cdot 2\cdot \frac{1}{3^{5/4}}$$

We compute $$\binom{9}{5} = \binom{9}{4} = 126$$. Also $$3^{5/4} = 3\cdot 3^{1/4} = 3\sqrt[4]{3}$$.

So $$T_6 = 126\cdot 2\cdot\frac{1}{3\sqrt[4]{3}} = \frac{252}{3\sqrt[4]{3}} = \frac{84}{\sqrt[4]{3}}$$.

We are told $$T_6 = \frac{\alpha}{\sqrt[4]{3}}$$, so $$\alpha = 84$$.

Hence, the correct answer is 84.