# The term independent of x in the expansion of

Question:

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 :

1. (1) $-72$

2. (2) 36

3. (3) $-36$

4. (4) $-108$

Correct Option: , 4

Solution:

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 $\left(2 x^{2}-\frac{3}{x^{2}}\right)^{0}$

$=\frac{-1}{60}{ }^{6} C_{3}(2)^{3}(3)^{3}+\frac{1}{81}{ }^{6} C_{5}(2)(3)^{5}$

$=-72+36=-36$