The term independent of $x$ 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
Given expression is,
$\left(\frac{1}{60}-\frac{x^{8}}{81}\right)\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$
$=\frac{1}{60}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{x^{8}}{81}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}$
Term independent of $x$,
$=$ Coefficient of $x^{\circ}$ in $\frac{1}{60}\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6}-\frac{1}{81}$
coefficient of $x^{8}$ in $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{6^{x^{2}}}$
coefficient of $x^{-8}$ in $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{0}$
$=\frac{-1}{60}{ }^{6} C_{3}(2)^{3}(3)^{3}+\frac{1}{81}{ }^{6} C_{5}(2)(3)^{5}$
$=-72+36=-36$