$The coefficient of the term independent of x in the expansion$$\text{ of }\left(1+x+2x^3\right){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9\text{ is }$

$\begin{array}{l}{\text{ The r}}^{\text{th }}\text{ term in the expansion of }{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9\text{ is }{T}_{r+1}{=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(-\frac{1}{3x}\right)}^r{=}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(-\frac{1}{3}\right)}^r{x}^{18-3r}\\ \text{ The coefficient of the term independent of x in the expansion of }\left(1+x+2x^3\right)\text{\hspace{0.17em} is }{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9\\ =\text{ Sum of the coefficient of the terms }{x}^0,x^{-1}\text{ and }{x}^{-3}\text{ in }\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)\\ \text{ Hence the term indepndent of }x\text{\hspace{0.17em}is }=1{\times }^9{C}_6{\left(\frac{3}{2}\right)}^{9-6}{\left(\frac{1}{3}\right)}^6+2{\times }^9{C}_7{\left(\frac{3}{2}\right)}^{9-7}{\left(\frac{1}{3}\right)}^7\\ =\frac{\mathrm{9.8.7}}{\mathrm{1.2.3}}\cdot \frac{3^3}{2^3}\cdot \frac{1}{3^6}+2\frac{9.8}{1.2}(-1)\frac{3^2}{2^2}\cdot \frac{1}{3^7}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}\end{array}$