Mathematics - Binomial Theorem Question with Solution | TestHub

$\begin{array}{rl}& \left(1+2x-3x^3\right){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9\\ & \text{ General term }m{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9\\ & ={}^9{C}_r\cdot \frac{3^{9-2r}}{2^{9-r}}(-1{)}^r\cdot x^{18-3r}\end{array}$ Put $r=6$ to get coeff. of $x^0={}^9{C}_6\cdot \frac{1}{6^3}\cdot x^0=\frac{7}{18}{x}^0$ Put $\mathrm{r}=7$ to get coeff. of ${\mathrm{x}}^{-3}={}^9{\mathrm{C}}_{\mathrm{r}}\cdot \frac{3^{-5}}{2^2}(-1{)}^7\cdot {\mathrm{x}}^{-3}$ $\begin{array}{rl}& =-{}^9{C}_7\cdot \frac{1}{3^5\cdot 2^2}\cdot x^{-3}=\frac{-1}{27}{x}^{-3}\\ & \left(1+2x-3x^3\right)\left(\frac{7}{18}{x}^0-\frac{1}{27}{x}^{-3}\right)\\ & \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2}\\ & \therefore 108\cdot \frac{1}{2}=54\end{array}$