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Question: The term independent of x in the expansion of \({\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\...

The term independent of x in the expansion of (x+1x23−x13+1−x−1x−x12)10{\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}} is
(a)120\left( a \right)120
(b)210\left( b \right)210
(c)310\left( c \right)310
(d)4\left( d \right)4

Explanation

Solution

In this particular question use the concept that according to Binomial theorem, the expansion of
(b−a)n{\left( {b - a} \right)^n} is equal to ∑r=0nnCrbn−r(−a)r\sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( { - a} \right)}^r}} so first simplify the given expression by writing (x+1)=((x13)3+1)\left( {x + 1} \right) = \left( {{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + 1} \right) and use the formula of (a3+b3)=(a+b)(a2−ab+b2)\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given expression:
(x+1x23−x13+1−x−1x−x12)10{\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}
Now the above expression is also written as
⇒((x13)3+1x23−x13+1−(x12)2−1x12(x12−1))10\Rightarrow {\left( {\dfrac{{{{\left( {{x^{\dfrac{1}{3}}}} \right)}^3} + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{{{\left( {{x^{\dfrac{1}{2}}}} \right)}^2} - 1}}{{{x^{\dfrac{1}{2}}}\left( {{x^{\dfrac{1}{2}}} - 1} \right)}}} \right)^{10}}
Now as we know that (a3+b3)=(a+b)(a2−ab+b2)\left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) and (a2−b2)=(a−b)(a+b)\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) so use these properties in the above equation we have,
⇒((x13+1)(x23−x13+1)x23−x13+1−(x12−1)(x12+1)x12(x12−1))10\Rightarrow {\left( {\dfrac{{\left( {{x^{\dfrac{1}{3}}} + 1} \right)\left( {{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1} \right)}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{\left( {{x^{\dfrac{1}{2}}} - 1} \right)\left( {{x^{\dfrac{1}{2}}} + 1} \right)}}{{{x^{\dfrac{1}{2}}}\left( {{x^{\dfrac{1}{2}}} - 1} \right)}}} \right)^{10}}
Now simplify it we have,
⇒(x13+1−(x12+1)x12)10\Rightarrow {\left( {{x^{\dfrac{1}{3}}} + 1 - \dfrac{{\left( {{x^{\dfrac{1}{2}}} + 1} \right)}}{{{x^{\dfrac{1}{2}}}}}} \right)^{10}}
⇒(x13+1−(1+x−12))10\Rightarrow {\left( {{x^{\dfrac{1}{3}}} + 1 - \left( {1 + {x^{\dfrac{{ - 1}}{2}}}} \right)} \right)^{10}}
⇒(x13+1−1−x−12)10\Rightarrow {\left( {{x^{\dfrac{1}{3}}} + 1 - 1 - {x^{\dfrac{{ - 1}}{2}}}} \right)^{10}}
⇒(x13−x−12)10\Rightarrow {\left( {{x^{\dfrac{1}{3}}} - {x^{\dfrac{{ - 1}}{2}}}} \right)^{10}}
Now we know according to Binomial theorem, the expansion of
(b−a)n=∑r=0nnCrbn−r(−a)r{\left( {b - a} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{b^{n - r}}{{\left( { - a} \right)}^r}}
So, on comparing b=x13, a=x−12, n=10b = {x^{\dfrac{1}{3}}},{\text{ }}a = {x^{\dfrac{{ - 1}}{2}}},{\text{ }}n = 10
⇒(x13−x−12)10=∑r=01010Cr(x13)10−r(−x−12)r\Rightarrow {\left( {{x^{\dfrac{1}{3}}} - {x^{\dfrac{{ - 1}}{2}}}} \right)^{10}} = \sum\limits_{r = 0}^{10} {{}^{10}{C_r}{{\left( {{x^{\dfrac{1}{3}}}} \right)}^{10 - r}}{{\left( { - {x^{\dfrac{{ - 1}}{2}}}} \right)}^r}}
⇒∑r=01010Cr(x)10−r3−r2(−1)r=∑r=01010Cr(x)20−2r−3r6(−1)r\Rightarrow \sum\limits_{r = 0}^{10} {{}^{10}{C_r}{{\left( x \right)}^{\dfrac{{10 - r}}{3} - \dfrac{r}{2}}}{{\left( { - 1} \right)}^r}} = \sum\limits_{r = 0}^{10} {{}^{10}{C_r}{{\left( x \right)}^{\dfrac{{20 - 2r - 3r}}{6}}}{{\left( { - 1} \right)}^r}}
Now, we want the term independent of xx
So, put the power of xx in the expansion of (x13−x−12)10{\left( {{x^{\dfrac{1}{3}}} - {x^{\dfrac{{ - 1}}{2}}}} \right)^{10}} equal to zero.
⇒20−2r−3r6=0\Rightarrow \dfrac{{20 - 2r - 3r}}{6} = 0
⇒5r=20\Rightarrow 5r = 20
⇒r=4\Rightarrow r = 4
So, put r=4,r = 4,in ∑r=01010Cr(x)20−2r−3r6(−1)r\sum\limits_{r = 0}^{10} {{}^{10}{C_r}{{\left( x \right)}^{\dfrac{{20 - 2r - 3r}}{6}}}{{\left( { - 1} \right)}^r}} we have
⇒∑r=01010Cr(x)20−2r−3r6(−1)r=10C4(x)0(−1)4\Rightarrow \sum\limits_{r = 0}^{10} {{}^{10}{C_r}{{\left( x \right)}^{\dfrac{{20 - 2r - 3r}}{6}}}{{\left( { - 1} \right)}^r}} = {}^{10}{C_4}{\left( x \right)^0}{\left( { - 1} \right)^4}
⇒10C4(1)\Rightarrow {}^{10}{C_4}\left( 1 \right)
Now as we know that nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} so use this property we have,
⇒10C4=10!4!(10−4)!=10!4!(6!)=10.9.8.7.6!(4.3.2.1).6!=10.9.8.74.3.2.1=210\Rightarrow {}^{10}{C_4} = \dfrac{{10!}}{{4!\left( {10 - 4} \right)!}} = \dfrac{{10!}}{{4!\left( {6!} \right)}} = \dfrac{{10.9.8.7.6!}}{{\left( {4.3.2.1} \right).6!}} = \dfrac{{10.9.8.7}}{{4.3.2.1}} = 210
So, this is the required term independent of xxin the expansion of (x+1x23−x13+1−x−1x−x12)10{\left( {\dfrac{{x + 1}}{{{x^{\dfrac{2}{3}}} - {x^{\dfrac{1}{3}}} + 1}} - \dfrac{{x - 1}}{{x - {x^{\dfrac{1}{2}}}}}} \right)^{10}}.
So this is the required answer.
Hence option (b) is the correct answer.

Note : Whenever we face such type of problem the key concept we have to remember is that always remember the general expansion of (b−a)n{\left( {b - a} \right)^n}, then in the expansion put the power of xx equal to zero, and calculate the value of rr, then put this value of rr in the expansion we will get the required term which is independent of xx.