Show that the term containing

For $\left(3 x-\frac{1}{2 x}\right)^{8}$

$a=3 x, b=\frac{-1}{2 x}$ and $n=8$

We have a formula,

$t_{r+1}=\left(\begin{array}{l}n \\ r\end{array}\right) a^{n-r} b^{r}$

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(3 x)^{8-r}\left(\frac{-1}{2 x}\right)^{r}$

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{8-r}(x)^{8-r}\left(\frac{-1}{2}\right)^{r}(x)^{-r}$

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{8-r}(x)^{8-r-r}\left(\frac{-1}{2}\right)^{r}$

$=\left(\begin{array}{l}8 \\ r\end{array}\right)(3)^{8-r}\left(\frac{-1}{2}\right)^{r}(x)^{8-2 r}$

To get coefficient of $x^{3}$ we must have,

$(x)^{8-2 r}=(x)^{3}$

$\cdot 8-2 r=3$

$\cdot 2 r=5$

$\cdot r=2.5$

As $\left(\begin{array}{l}8 \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}8 \\ 2.5\end{array}\right)$ is not possible

Therefore, the term containing $x^{3}$ does not exist in the expansion of $\left(3 x-\frac{1}{2 x}\right)^{8}$