If $p$ and $q$ are respectively the coefficient of $x^{-3}$ and $x^{-5}$ in the expansion of
${\left(x^{1/3}+\frac{1}{2x^{1/3}}\right)}^{21}, x>0,$ then $\frac{5p}{4q}=$

$E={\left(x^{\frac{1}{3}}+\frac{1}{2x^{{\displaystyle \frac{1}{3}}}}\right)}^{21}$, $x>0$

$\bullet p=$ Coefficient of $x^{-3}$ in $E$

$\Rightarrow p={}^{21}C_r{\left(x^{\frac{1}{3}}\right)}^{21-r}{\left(\frac{1}{2}{x}^{-\frac{1}{3}}\right)}^r$, where $T_{r+1}$ is the corresponding term

So, $\frac{21-r}{3}-\frac{r}{3}=-3$

$\Rightarrow 21-2r=-9$

$\Rightarrow r=15$

So, $p=\left({}^{21}C_{15}\right){\left(\frac{1}{2}\right)}^{15}$

$\bullet q=$ Coefficient of $x^{-5}$ in $E$

$\Rightarrow q={}^{21}C_3{\left(x^{\frac{1}{3}}\right)}^{21-r}{\left(\frac{1}{2}{x}^{-\frac{1}{3}}\right)}^r$, where $T_{r+1}$ is the corresponding terms

So, $\frac{21-r}{3}-\frac{r}{3}=-5$

$\Rightarrow 2r=36$

$\Rightarrow r=18$

So, $q=\left({}^{21}C_{18}\right){\left(\frac{1}{2}\right)}^{18}$

Now, $\frac{5p}{4q}=\frac{{}^{21}C_{15}{\left({\displaystyle \frac{1}{2}}\right)}^{15}}{{}^{21}C_{18}{\left({\displaystyle \frac{1}{2}}\right)}^{18}}=\left(\frac{{}^{21}C_6}{{}^{21}C_3}\right)8$

$\Rightarrow \frac{5p}{4q}=\left(\frac{21!}{6!15!}\right)\left(\frac{3!18!}{21!}\right)\left(8\right)=408$