If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of $\left(2^{1/3}+\right.$${\left.3^{-1/3}\right)}^n$ is $1/6$ then the value of $n$is

Let  $a=7^{\text{th }}$term from the beginning

$=T_7=T_{6+1}{=}^n{C}_6{\left(2^{1/3}\right)}^{n-6}{\left(3^{-1/3}\right)}^6$

and let $b=7^{\text{th }}$term from the end

$=(n+2-7)\text{ th or }(n-5)\text{ th }$ term from the beginning

$\begin{array}{l}{=}^n{C}_{n-6}{\left(2^{1/3}\right)}^{n-(n-6)}{\left(3^{-1/3}\right)}^{n-6}\\ {=}^n{C}_6{\left(2^{1/3}\right)}^6{\left(3^{-1/3}\right)}^{n-6}\end{array}$

we are given 

$\begin{array}{l}\frac{a}{b}=\frac{1}{6}\\ \frac{{\left(2^{1/3}\right)}^{n-6}{\left(3^{-1/3}\right)}^6}{{\left(2^{1/3}\right)}^6{\left(3^{-1/3}\right)}^{n-6}}=\frac{1}{6}\\ \Rightarrow {\left(\frac{\left(2^{1/3}\right)}{\left(3^{-1/3}\right)}\right)}^{n-12}=\frac{1}{6}\\ \Rightarrow 6^{(n-12)/3}=6^{-1}\\ \Rightarrow n-12=-3\Rightarrow n=9.\end{array}$