# Find n in the binomial

Question:

Find $n$ in the binomial $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, if the ratio of $7^{\text {th }}$ term from the beginning to the $7^{\text {th }}$ term from the end is $\frac{1}{6}$.

Solution:

Given expression is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$

Now, $7^{\text {th }}$ term from beginning, $T_{7}=T_{6+1}={ }^{n} C_{6}(\sqrt[3]{2})^{n-6}\left(\frac{1}{\sqrt[3]{3}}\right)^{6}$

And $7^{\text {th }}$ term from end is same as $7^{\text {th }}$ term from the beginning of $\left(\frac{1}{\sqrt[3]{3}}+\sqrt[3]{2}\right)^{n}$

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

Given that

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

$\frac{(\sqrt[3]{2})^{n-12}}{\left(\frac{1}{\sqrt[3]{3}}\right)^{n-12}}=\frac{1}{6} \Rightarrow(\sqrt[3]{2} \sqrt[3]{3})^{n-12}=6^{-1} \Rightarrow 6^{\frac{n-12}{3}}=6^{-1}$

$\frac{n-12}{3}=-1 \quad \Rightarrow \quad n=9$