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

Hint: 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,
\[\begin{align}
  & \Rightarrow \left( \dfrac{1}{60}-\dfrac{{{x}^{8}}}{81} \right).{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}} \\
 & \Rightarrow \dfrac{1}{60}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}-\dfrac{{{x}^{8}}}{81}{{\left( 2{{x}^{2}}-\dfrac{3}{{{x}^{2}}} \right)}^{6}}...............\left( 1 \right) \\
\end{align}\]
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
\[\begin{align}
  & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}{{\left( 2{{x}^{2}} \right)}^{r-1}}{{\left( \dfrac{-3}{{{x}^{2}}} \right)}^{6-r+1}} \\
 & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}\dfrac{1}{{{\left( {{x}^{2}} \right)}^{7-r}}} \\
 & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}{{\left( {{x}^{2}} \right)}^{r-7}} \\
 & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{x}^{2}}^{r-2}{{\left( -3 \right)}^{6-r+1}}{{x}^{2r-14}} \\
 & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{r-1}}\times {{2}^{r-1}}{{\left( -3 \right)}^{6-r+1}}{{x}^{4r-16}} \\
\end{align}\]
As we need the term independent of x,
$\begin{align}
  & \Rightarrow 4r-16=0 \\
 & \Rightarrow 4r=16 \\
 & \Rightarrow r=4 \\
\end{align}$
So, we get that ${{4}^{th}}$ term is independent of x. Then value of ${{4}^{th}}$ term in the part 1 is
\[\begin{align}
  & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{4-1}}\times {{2}^{4-1}}{{\left( -3 \right)}^{6-4+1}}{{x}^{4\left( 4 \right)-16}} \\
 & \Rightarrow \dfrac{1}{60}\times {}^{6}{{C}_{3}}\times {{2}^{3}}{{\left( -3 \right)}^{3}}{{x}^{0}} \\
 & \Rightarrow \dfrac{1}{60}\times \dfrac{6!}{3!\times 3!}\times 8\times \left( -27 \right) \\
 & \Rightarrow \dfrac{1}{60}\times \dfrac{720}{6\times 6}\times 8\times \left( -27 \right) \\
 & \Rightarrow \dfrac{1}{60}\times 20\times 8\times \left( -27 \right) \\
 & \Rightarrow \dfrac{1}{60}\times \left( -4320 \right) \\
 & \Rightarrow -72 \\
\end{align}\]
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.