Find n in the binomial

In the binomail expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n},[(n+1)-7+1]^{\text {th }}$ i.e., $(n-5)^{\text {th }}$ term from the beginning is the $7^{\text {th }}$ term from the end.

Now,

$T_{7}={ }^{n} C_{6}(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^{0}={ }^{n} C_{6} \times 2^{\frac{n}{3}-2} \times \frac{1}{3^{2}}$

And,

$T_{n-5}={ }^{n} C_{n-6}(\sqrt[3]{2})^{6}\left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}={ }^{n} C_{6} \times 2^{2} \times \frac{1}{3^{\frac{n}{3}-2}}$

It is given that,

$\frac{T_{7}}{T_{n-5}}=\frac{1}{6}$

$\Rightarrow \frac{{ }^{n} C_{6} \times 2^{\frac{n}{3}-2} \times \frac{1}{3^{2}}}{{ }^{n} C_{6} \times 2^{2} \times \frac{1}{3^{\frac{n}{3}-2}}}=\frac{1}{6}$

$\Rightarrow 2^{\frac{n}{3}-2-2} \times 3^{\frac{n}{3}-2-2}=\frac{1}{6}$

$\Rightarrow\left(\frac{1}{6}\right)^{4-\frac{n}{3}}=\frac{1}{6}$

$\Rightarrow 4-\frac{n}{3}=1$

$\Rightarrow n=9$

Hence, the value of n is 9.