# Find $n$ in the binomial $\big[\sqrt[3]{2} +\large\frac{1}{\sqrt[3]{3}} \big]^n$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\large\frac{1}{6}$

$\begin{array}{1 1}(A)\;9\\(B)\;6\\(C)\;12\\(D)\;\text{None of these}\end{array}$

Toolbox:
• $T_{r+1}=nC_ra^{n-r}b^r$
$T_7$ in $\big[\sqrt[3]{2} +\large\frac{1}{\sqrt[3]{3}} \big]^n=$$nC_6(2^{1/3})^{n-6}(\large\frac{1}{3^{1/3}})^6 7^{th} term from the end in \big[\sqrt[3]{2} +\large\frac{1}{\sqrt[3]{3}} \big]^n \Rightarrow T_7 in \big[\sqrt[3]{2} +\large\frac{1}{\sqrt[3]{3}} \big]^n \Rightarrow nC_6(\large\frac{1}{3^{1/3}})^{n-6}$$(2^{1/3})^6$
$\therefore \large\frac{nC_6(2^{1/2})^{n-6}(\large\frac{1}{3^{1/6}})^6}{nC_6(\large\frac{1}{3^{1/3}})^{n-6}(2^{1/3})^6}=\frac{1}{6}$
$\Rightarrow \large\frac{(2^{\Large\frac{1}{3}})^{n+2}}{(\large\frac{1}{3^{1/3}})^{n-12}}=\frac{1}{6}$
$\Rightarrow 2^{\Large\frac{n-12}{3}}.3^{\Large\frac{n-12}{3}}=\large\frac{1}{6}$
$\Rightarrow 6^{\large\frac{n-12}{3}}=6^{-1}$