Question

The coefficient of the term independent of $x$ in the expansion of ${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$ is

• A. $194$
• B. $180$
• C. $210$
• D. $245$

Answer 1

The correct option is C $210$

Given: ${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$
Now, $\frac{(x^{1/3}{)}^3+1}{x^{2/3}-x^{1/3}+1}=\frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1}$
$\Rightarrow \frac{x+1}{x^{2/3}-x^{1/3}+1}=x^{1/3}+1$

And $\frac{x-1}{x-\sqrt{x}}=\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}$
$\Rightarrow \frac{x-1}{x-\sqrt{x}}=1+\frac{1}{\sqrt{x}}$

So, the given expression becomes
${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$
$={[(1+x^{1/3})-(1+\frac{1}{\sqrt{x}})]}^{10}$
$={(x^{1/3}-\frac{1}{\sqrt{x}})}^{10}$

General term,
$T_{r+1}={}^{10}{C}_r\cdot {\left(x^{1/3}\right)}^{10-r}\cdot {\left(\frac{-1}{\sqrt{x}}\right)}^r$
$=(-1{)}^r\cdot {}^{10}{C}_r\cdot (x{)}^{\frac{10-r}{3}-\frac{r}{2}}$

For term to be independent of $x,$
$\frac{10-r}{3}-\frac{r}{2}=0$
$\Rightarrow 20-2r=3r$
$\Rightarrow r=4$
$\therefore$ Term independent of $x$ is
$T_{4+1}=(-1{)}^4\cdot {}^{10}{C}_4=210$