The term independent of x in the expansion of160−x881⋅2×2−3×26 is equal to

# The term independent of $x$ in the expansion ofis equal to

1. A

-108

2. B

-72

3. C

-36

4. D

36

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### Solution:

Given expansion is $\left(\frac{1}{60}-\frac{{x}^{8}}{81}\right){\left(2{x}^{2}-\frac{3}{{x}^{2}}\right)}^{6}$

$=\frac{1}{60}{\left(2{x}^{2}-\frac{3}{{x}^{2}}\right)}^{6}-\frac{{x}^{8}}{81}{\left(2{x}^{2}-\frac{3}{{x}^{2}}\right)}^{6}$

So its general term is

$\begin{array}{l}{T}_{r+1}={\frac{1}{60}}^{6}{C}_{r}{\left(2{x}^{2}\right)}^{6-r}{\left(\frac{-3}{{x}^{2}}\right)}^{r}-\frac{{x}^{8}}{81}{\cdot }^{6}{C}_{r}{\left(2{x}^{2}\right)}^{6-r}{\left(\frac{-3}{{x}^{2}}\right)}^{r}\\ ={\frac{1}{60}}^{6}{C}_{r}\left(2{\right)}^{6-r}\left(-3{\right)}^{r}{x}^{12-2r-2r}-{\frac{1}{81}}^{6}{C}_{r}\left(2{\right)}^{6-r}\left(-3{\right)}^{r}{x}^{12-2r-2r+8}\\ ={\frac{1}{60}}^{6}{C}_{r}\left(2{\right)}^{6-r}\left(-3{\right)}^{r}{x}^{12-4r}-{\frac{1}{81}}^{6}{C}_{r}\left(2{\right)}^{6-r}\left(-3{\right)}^{r}{x}^{20-4r}\end{array}$

For this term to be independent of $x,$put $r=3$ in

1st part and r = 5 in 2nd part.

From ,

$=-72+36=-36$

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