$\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$ ના વિસ્તરણનું અચળ પદ શોધો. 

We have ${T_{r + 1}} = {\,^6}{C_r}{\left( {\frac{3}{2}{x^2}} \right)^{6 - r}}\left( { - \frac{1}{{3x}}} \right)$

$ = {\,^6}{C_r}{\left( {\frac{3}{2}} \right)^{6 - r}}{\left( {{x^2}} \right)^{6 - r}}{( - 1)^r}{\left( {\frac{1}{x}} \right)^r}\left( {\frac{1}{{{3^r}}}} \right)$

$ = {( - 1)^r}{\quad ^6}{C_r}\quad \frac{{{{(3)}^{6 - 2r}}}}{{{{(2)}^{6 - r}}}}\quad {x^{12 - 3r}}$

The term will be independent of $x$ if the index of $x$ is zero, i.e., $12-3 r=0 .$ Thus, $r=4$

Hence $5^{\text {th }}$ term is independent of $x$ and is given by ${( - 1)^4}{\,^6}{C_4}\frac{{{{(3)}^{6 - 8}}}}{{{{(2)}^{6 - 4}}}} = \frac{5}{{12}}$