Question

The term independent of x in the expansion of $\left( \dfrac{1}{60}-\dfrac{{{x}^{8}}}{81} \right).{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$ is equal to
(A) 36
(B) -108
(C) -72
(D) -36

Answer 1

We solve this question by dividing the given expression into two parts and then finding the term that is independent of x in each part and then finding the value of that term in the expansion using the formula for binomial expansion ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+........+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. Then subtract the obtained result in the given order to get the final result.

Complete step-by-step answer :
We are given, $\left( \dfrac{1}{60}-\dfrac{{{x}^{8}}}{81} \right).{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$
It can be simplified as,

& \Rightarrow \left( \dfrac{1}{60}-\dfrac{{{x}^{8}}}{81} \right).{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}} \\\ & \Rightarrow \dfrac{1}{60}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}-\dfrac{{{x}^{8}}}{81}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}...............\left( 1 \right) \\\ \end{aligned}$$ Let us divide the expression into two parts such that $$\dfrac{1}{60}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$ and $$\dfrac{{{x}^{8}}}{81}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$. Let us try to find the term independent of x from part 1. Now let us consider the formula for binomial expansion, $${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+........+{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$$ The value of ${{r}^{th}}$ term in the expansion is $${}^{n}{{C}_{r-1}}{{a}^{r-1}}{{b}^{n-r+1}}$$ Now, let us apply it on the first part, that is $$\dfrac{1}{60}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$. Then the ${{r}^{th}}$ term in the expansion can be given by $$\begin{aligned} & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}{{\left( 2{{x}^{2}} \right)}^{r-1}}{{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{6-r+1}} \\\ & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}\dfrac{1}{{{\left( {{x}^{2}} \right)}^{7-r}}} \\\ & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}{{\left( {{x}^{2}} \right)}^{r-7}} \\\ & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}{{x}^{2r-14}} \\\ & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{\left( -3 \right)}^{6-r+1}}{{x}^{4r-16}} \\\ \end{aligned}$$ As we need the term independent of x, $\begin{aligned} & \Rightarrow 4r-16=0 \\\ & \Rightarrow 4r=16 \\\ & \Rightarrow r=4 \\\ \end{aligned}$ So, we get that ${{4}^{th}}$ term is independent of x. Then value of ${{4}^{th}}$ term in the part 1 is $$\begin{aligned} & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{4-1}}\times {{2}^{4-1}}{{\left( -3 \right)}^{6-4+1}}{{x}^{4\left( 4 \right)-16}} \\\ & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{3}}\times {{2}^{3}}{{\left( -3 \right)}^{3}}{{x}^{0}} \\\ & \Rightarrow \dfrac{1}{60}\times \dfrac{6!}{3!\times 3!}\times 8\times \left( -27 \right) \\\ & \Rightarrow \dfrac{1}{60}\times \dfrac{720}{6\times 6}\times 8\times \left( -27 \right) \\\ & \Rightarrow \dfrac{1}{60}\times 20\times 8\times \left( -27 \right) \\\ & \Rightarrow \dfrac{1}{60}\times \left( -4320 \right) \\\ & \Rightarrow -72 \\\ \end{aligned}$$ So constant from part 1 is -72. From part 2, we have $$\dfrac{{{x}^{8}}}{81}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$. For the term in it to be constant we need to find the terms in the expansion $${{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$ which results in ${{x}^{-8}}$. The (r+1) term in the expansion will be $${{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$. $${}^{6}{{C}_{r}}{{\left( 2{{x}^{2}} \right)}^{r}}{{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{6-r}}$$ $$\Rightarrow $$$${}^{6}{{C}_{r}}{{2}^{r}}{{(-3)}^{6-r}}\left( \dfrac{{{x}^{2r}}}{{{x}^{2(6-r)}}} \right)$$ $$\Rightarrow $$$${}^{6}{{C}_{r}}{{2}^{r}}{{(-3)}^{6-r}}\left( {{x}^{2r-12+2r}} \right)$$ $$\Rightarrow $$$${}^{6}{{C}_{r}}{{2}^{r}}{{(-3)}^{6-r}}\left( {{x}^{4r-12}} \right)$$ So, from this we can say that $$4r-12=-8$$ $$4r=12-8$$ $$\Rightarrow 4r=4$$ $$\Rightarrow r=1$$ Substituting r=1 in the value of expansion we get, $$\Rightarrow {}^{6}{{C}_{1}}{{\left( 2{{x}^{2}} \right)}^{1}}{{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{6-1}}$$ $$\Rightarrow {}^{6}{{C}_{1}}{{\left( 2{{x}^{2}} \right)}^{1}}{{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{5}}$$ $$\Rightarrow 6\times 2\times {{(-3)}^{5}}\left( \dfrac{{{x}^{2}}}{{{x}^{10}}} \right)$$ $$\Rightarrow 6\times 2\times {{(-3)}^{5}}\left( \dfrac{1}{{{x}^{8}}} \right)$$ As we need the value of $$\dfrac{{{x}^{8}}}{81}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}$$, multiplying with the term $$\dfrac{{{x}^{8}}}{81}$$, we get $$\Rightarrow 6\times 2\times {{(-3)}^{5}}\left( \dfrac{1}{{{x}^{8}}} \right)\left( \dfrac{{{x}^{8}}}{81} \right)$$ $$\Rightarrow \dfrac{6\times 2\times (81)\times (-3)}{81}$$$$=-36$$ Substituting the values in (1) we get $$(-72)-(-36)=-72+36=-36$$ So, the term independent of x is -36. **Hence answer is option D.** **Note** : The major mistake one does while solving this problem is one might forget to multiply the obtained result with $$\dfrac{1}{60}$$ in the first part and with $$\dfrac{1}{81}$$, in the second part. So, one needs to look at the terms carefully and multiply them at the end.