Answer:
Option C
Explanation:
$\therefore$ $\left[\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right]^{10} $
= $\left[\frac{(x^{\frac{1}{3}})^{3}+1^{3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} $
= $\left[\frac{(x^{\frac{1}{3}}+1)(x^{2/3}+1-x^{1/3}}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{((\sqrt{x}))^{2}-1}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} $
= $\left[(x^{1/3}+1)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right]^{10}$
= $(x^{1/3}-x^{-1/2})^{10}$
$ \therefore$ The general term is
$T_{r+1}=^{10}C_{r}(x^{1/3})^{10-r}(-x^{-1/2})^{r}$
= $^{10}C_{r}(-1)^{r}x^{\frac{10-r}{3}-\frac{r}{2}}$
For independent of x, put
$\frac{10-r}{3}-\frac{r}{2}=0\Rightarrow 20-2r-3r=0$
$\Rightarrow$ $ 20=5r\Rightarrow r=4$
$\therefore$ $T_{5}=^{10}C_{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$
