Question:
$\mathrm{x}, \mathrm{x} \neq 0$, in the expansion of $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$.
Solution:
Given $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$
From the standard formula of $T_{r+1}$ we can write given expression as
$T_{r+1}={ }^{15} C_{r}\left(\frac{3 x^{2}}{2}\right)^{15-r}\left(-\frac{1}{3 x}\right)^{r}$
$T_{r+1}={ }^{15} C_{r}(-1)^{r} 3^{15-2 r} 2^{r-15} x^{30-3 r}$
For the term independent of $x$, we have
$30-3 r=0$
Which implies $r=10$
By substituting the value of $r$ in above obtained expression we get
$T_{10+1}={ }^{15} C_{10} 3^{-5} 2^{-5}$
$={ }^{15} C_{10}\left(\frac{1}{6}\right)^{5}$