A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${\left(2^{\frac{1}{3}}+\frac{1}{2(3{)}^{\frac{1}{3}}}\right)}^{10} \mathrm{is}$

Since, rth term from the end in the expansion of a binomial (r +$\mathrm{\alpha }$)ⁿ is same as the (n - r +2)th term from the beginning in the expansion of same binomial.

$\begin{array}{r}\therefore \text{ Required ratio }=\frac{{\mathrm{T}}_5}{{\mathrm{T}}_{10-5+2}}\\ =\frac{{\mathrm{T}}_5}{{\mathrm{T}}_7}=\frac{{\mathrm{T}}_{4+1}}{{\mathrm{T}}_{6+1}}\end{array}$

$\Rightarrow \frac{{\mathrm{T}}_5}{{\mathrm{T}}_{10-5+2}}=\frac{{ }^{10}{\mathrm{C}}_4{\left(2^{1/3}\right)}^{10-4}{\left(\frac{1}{2(3{)}^{1/3}}\right)}^4}{{ }^{10}{\mathrm{C}}_6{\left(2^{1/3}\right)}^{10-6}{\left(\frac{1}{2(3{)}^{1/3}}\right)}^6}$

$\begin{array}{r}\left[\because {\mathrm{T}}_{\mathrm{r}+1}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}{\mathrm{a}}^{\mathrm{r}}\right]\\ \left[{\because }^{10}{\mathrm{C}}_4{=}^{10}{\mathrm{C}}_6\right]\end{array}$

$\begin{array}{r}=\frac{2^{6/3}{\left(2(3{)}^{1/3}\right)}^6}{2^{4/3}{\left(2(3{)}^{1/3}\right)}^4}\\ =2^{6/3-4/3}{\left(2(3{)}^{1/3}\right)}^{6-4}\\ =2^{23}\cdot 2^2\cdot 3^{23}\\ =4(6{)}^{23}\\ =4(36{)}^{1/3}\end{array}$

So, the required ratio is. $4(36{)}^{1/3}:1$