Find the term independent of

Given

$\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$

Consider

$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$

Using standard formula above expression can be written as

$T_{r+1}={ }^{9} C_{r}\left(\frac{3}{2} x^{2}\right)^{9-r}\left(-\frac{1}{3 x}\right)^{r}={ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}$

Hence the general term in the expression of given expansion is

${ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{18-3 r}+{ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{19-3 r}$

$+2 \cdot{ }^{9} C_{r}\left(\frac{3}{2}\right)^{9-r}\left(-\frac{1}{3}\right)^{r} x^{21-3 r}$

For independent term of $x$, substitute $18-3 r=0$ $19-3 r=0$ and $21-3 r=0$

We get $r=6$ and $r=7$

Hence second term is not independent of $x$ Therefore, term independent of $\mathrm{x}$ is

${ }^{9} C_{6}\left(\frac{3}{2}\right)^{9-6}\left(-\frac{1}{3}\right)^{6}+2 \cdot{ }^{9} C_{7}\left(\frac{3}{2}\right)^{9-7}\left(-\frac{1}{3}\right)^{7}$

$=\frac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2} \cdot \frac{1}{2^{3} \cdot 3^{3}}-2 \cdot \frac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1} \cdot \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}}$

$=\frac{84}{8} \cdot \frac{1}{3^{3}}-\frac{36}{4} \cdot \frac{2}{3^{5}}=\frac{21-4}{54}=\frac{17}{54}$