Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$, 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 _________.

<p>Fifth term from beginning $$ = {}^n{C_4}{\left( {{2^{{1 \over 4}}}} \right)^{n - 4}}{\left( {{3^{{{ - 1} \over 4}}}} \right)^4}$$</p> <p>Fifth term from end $= {(n - 5 + 1)^{th}}$ term from begin $$ = {}^n{C_{n - 4}}{\left( {{2^{{1 \over 4}}}} \right)^3}{\left( {{3^{{{ - 1} \over 4}}}} \right)^{n - 4}}$$</p> <p>Given $${{{}^n{C_4}{2^{{{n - 4} \over 4}}}\,.\,{3^{ - 1}}} \over {{}^n{C_{n - 3}}{2^{{4 \over 4}}}\,.\,{3^{ - \left( {{{n - 4} \over 4}} \right)}}}} = {6^{{1 \over 4}}}$$</p> <p>$\Rightarrow {6^{{{n - 8} \over 4}}} = {6^{{1 \over 4}}}$</p> <p>$\Rightarrow {{n - 8} \over 4} = {1 \over 4} \Rightarrow n = 9$</p> <p>$${T_6} = {T_{5 + 1}} = {}^9{C_5}{\left( {{2^{{1 \over 4}}}} \right)^4}{\left( {{3^{{{ - 1} \over 4}}}} \right)^5}$$</p> <p>$$ = {{{}^9{C_5}\,.\,2} \over {{3^{{1 \over 4}}}\,.\,3}} = {{84} \over {{3^{{1 \over 4}}}}} = {\alpha \over {{3^{{1 \over 4}}}}}$$</p> <p>$\Rightarrow \alpha = 84.$</p>