Question:
The term independent of $\mathrm{x}$ in the expansion of
$\left(\frac{1}{60}-\frac{x^{8}}{81}\right) \cdot\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$ is equal to :
Correct Option: , 4
Solution:
$\frac{1}{60}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{1}{81} \cdot x^{8}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$
its general term
$\frac{1}{60}{ }^{6} \mathrm{C}_{\mathrm{r}} 2^{6-\mathrm{r}}(-3)^{\mathrm{r}} \mathrm{x}^{12-\mathrm{r}}-\frac{1}{81}{ }^{6} \mathrm{C}_{\mathrm{r}} 2^{6-\mathrm{r}}(-3)^{\mathrm{r}} 12^{20-4 \mathrm{r}}$
for term independent of $x, r$ for $I^{\text {st }}$ expression is 3 and $r$ for second expression is 5
$\therefore$ term independent of $\mathrm{x}=-36$