If the last term in the binomial expansion of ${\left(2^{1/3}-\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}} \mathrm{is} {\left(\frac{1}{3^{5/3}}\right)}^{{\log }_{3}8}$, the the 5th term from  the begging is  

 

Solution:

$\mathrm{Last} \mathrm{term} \mathrm{of} {\left(2^{1/3}-\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}} \mathrm{is}$

$\begin{array}{r}{\mathrm{T}}_{\mathrm{n}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}{\left(2^{1/3}\right)}^{\mathrm{n}-\mathrm{n}}{\left(-\frac{1}{\sqrt{2}}\right)}^{\mathrm{n}}\\ {=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}(-1{)}^{\mathrm{n}}\frac{1}{2^{\mathrm{n}/2}}=\frac{(-1{)}^{\mathrm{n}}}{2^{\mathrm{n}/2}}\end{array}$

Also, we have

${\left(\frac{1}{3^{5/3}}\right)}^{{\log }_{3}8}=3^{-(5/3){\log }_3{2}^3}=2^{-5}$

$\begin{array}{r}\text{ Thus, }\frac{(-1{)}^{\mathrm{n}}}{2^{\mathrm{n}/2}}=2^{-5}\\ \Rightarrow \frac{(-1{)}^{\mathrm{n}}}{2^{\mathrm{n}/2}}=\frac{(-1{)}^{10}}{2^5}\\ \Rightarrow \frac{\mathrm{n}}{2}=5\Rightarrow \mathrm{n}=10\end{array}$

Now $\begin{array}{r}{\mathrm{T}}_5={\mathrm{T}}_{4+1}{=}^{10}{\mathrm{C}}_4{\left(2^{1/3}\right)}^{10-4}{\left(-\frac{1}{\sqrt{2}}\right)}^4\\ =\frac{10!}{4!6!}{\left(2^{1/3}\right)}^6(-1{)}^4{\left(2^{-1/2}\right)}^4\\ =210\left(2{)}^2\right(1)\left(2^{-2}\right)=210\end{array}$