Question

The term independent of ‘x’ (x > 0, $x\ne 0$) in the expansion of
${{\left[ \dfrac{\left( x+1 \right)}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( x-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$ is,
A. 105
B. 210
C. 315
D. 420

Answer 1

Hint: By using the formula ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$ make the first term in the form of ${{a}^{3}}+{{b}^{3}}$ and the second term in the form of ${{a}^{2}}-{{b}^{2}}$ and then use the expansion formula as ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Then use the formula ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}\times {{b}^{r}}$ of general term in the expansion and check at what value of ‘r’ it is becoming independent of x, and then put that value in the general term formula to get the final answer.

Complete step-by-step answer:
To solve the given question we will write down the given equation first and assume it as ‘S’, therefore,
$L={{\left[ \dfrac{\left( x+1 \right)}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( x-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
Above equation can also be written as,
$\Rightarrow L={{\left[ \dfrac{\left( {{x}^{\dfrac{3}{3}}}+{{1}^{3}} \right)}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( {{x}^{\dfrac{2}{2}}}-{{1}^{2}} \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
To proceed further in the solution we should know the formula given below,
Formula:
${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$
By using above formula we can write ‘S’ as,
$\Rightarrow L={{\left[ \dfrac{\left( {{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}+{{1}^{3}} \right)}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( {{\left( {{x}^{\dfrac{1}{2}}} \right)}^{2}}-{{1}^{2}} \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
Now to proceed further in the solution we should know the formulae given below,
Formula:
${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
By using the formulae given above in ‘L’ we will get,
$\Rightarrow L={{\left[ \dfrac{\left( {{x}^{\dfrac{1}{3}}}+1 \right)\left[ {{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}-{{x}^{\dfrac{1}{3}}}\times 1+{{1}^{2}} \right]}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( {{x}^{\dfrac{1}{2}}}+1 \right)\left( {{x}^{\dfrac{1}{2}}}-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
By using the formula ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$ and simplifying above equation further we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\dfrac{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
As the term ${{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1$ is present in both the numerator and denominator of the above equation therefore we can cancel it out, therefore we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\dfrac{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$
If we take $\sqrt{x}$ common from the denominator of the second term we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\dfrac{\left( \sqrt{x}+1 \right)\left( \sqrt{x}-1 \right)}{\sqrt{x}\left( \sqrt{x}-1 \right)} \right]}^{10}}$
As the term $\left( \sqrt{x}-1 \right)$ is present in both the numerator and denominator of the above equation therefore we can cancel it out, therefore we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\dfrac{\left( \sqrt{x}+1 \right)}{\sqrt{x}} \right]}^{10}}$
If we separate the denominator of the second term of the above equation we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\left( \dfrac{\sqrt{x}}{\sqrt{x}}+\dfrac{1}{\sqrt{x}} \right) \right]}^{10}}$
As the term $\sqrt{x}$ is present in both the numerator and denominator of the above equation therefore we can cancel it out, therefore we will get,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\left( 1+\dfrac{1}{{{x}^{\dfrac{1}{2}}}} \right) \right]}^{10}}$
As we know $\dfrac{1}{a}$ can also be written as ${{a}^{-1}}$ therefore above equation will become,
$\Rightarrow L={{\left[ \left( {{x}^{\dfrac{1}{3}}}+1 \right)-\left( 1+{{x}^{-\dfrac{1}{2}}} \right) \right]}^{10}}$
If we open the brackets in the above equation we will get,
$\Rightarrow L={{\left[ {{x}^{\dfrac{1}{3}}}+1-1-{{x}^{-\dfrac{1}{2}}} \right]}^{10}}$
By simplifying the above equation we will get,
$\Rightarrow L={{\left[ {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right]}^{10}}$ …………………………………………….. (1)
Now to proceed further in the solution we should know the formula of general term of the binomial expansion given below,
Formula:
The general term of ${{\left( a+b \right)}^{n}}$ is given by ${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}\times {{b}^{r}}$
If we observe equation (1) and the above equation then we can say that n = 10, and by using n = 10 we can write the general term of the expansion as,
${{T}_{r+1}}={}^{10}{{C}_{r}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-r}}\times {{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{r}}$
Now if we use the formula ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the above equation we will get,
${{T}_{r+1}}={}^{10}{{C}_{r}}\times {{x}^{\dfrac{10-r}{3}}}\times {{x}^{-\dfrac{r}{2}}}$
If we use the formula ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ in the above equation we will get,
$\Rightarrow {{T}_{r+1}}={}^{10}{{C}_{r}}\times {{x}^{\left( \dfrac{10-r}{3}-\dfrac{r}{2} \right)}}$ …………………………………………………. (2)
As we have to find the term of the expansion independent of ‘x’ and in the above general term the value of x must be 1.
As we know that ${{x}^{0}}=1$ therefore the power of ‘x’ in the general term should be zero to get the term independent of ‘x’ therefore we will get,
$\Rightarrow \dfrac{10-r}{3}-\dfrac{r}{2}=0$
$\Rightarrow \dfrac{2\left( 10-r \right)-3r}{3\times 2}=0$
$\Rightarrow \dfrac{20-2r-3r}{6}=0$
$\Rightarrow 20-5r=0\times 6$
$\Rightarrow 20-5r=0$
$\Rightarrow 20=5r$
$\Rightarrow \dfrac{20}{5}=r$
By rearranging and simplifying the above equation we will get,
Therefore, r = 4
Therefore if we put the value of ‘r’ in equation (2) we will get x = 1 and the term will become independent of ‘x’ therefore we will get,
$\Rightarrow {{T}_{4+1}}={}^{10}{{C}_{4}}\times 1$
By using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ in the above equation we will get,
$\Rightarrow {{T}_{5}}=\dfrac{10!}{4!\left( 10-4 \right)!}$
Further simplification in the above equation will give,
$\Rightarrow {{T}_{5}}=\dfrac{10!}{4!\times 6!}$
$\Rightarrow {{T}_{5}}=\dfrac{10\times 9\times 8\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}$
$\Rightarrow {{T}_{5}}=\dfrac{10\times 9\times 8\times 7}{8\times 3}$
$\Rightarrow {{T}_{5}}=\dfrac{10\times 9\times 7}{3}$
$\Rightarrow {{T}_{5}}=10\times 3\times 7$
$\Rightarrow {{T}_{5}}=210$
Therefore the 5th term of the expansion of ${{\left[ \dfrac{\left( x+1 \right)}{\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}-\dfrac{\left( x-1 \right)}{\left( x-\sqrt{x} \right)} \right]}^{10}}$ is independent of ‘x’ and it’s value is equal to 210.
Therefore the correct answer is option (b).

Note: When you get the value of r = 1 after do crosscheck it by substituting this value in ${{x}^{\left( \dfrac{10-r}{3}-\dfrac{r}{2} \right)}}$ and check whether it’s value is 1 or not.