Question

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

• A. $210$
• B. $105$
• C. $70$
• D. $112$

Answer 1

The correct option is A $210$

Given expression ${[\frac{{\left(x^{1/3}\right)}^3+1}{x^{2/3}-x^{1/3}+1}-\frac{{\left(x^{1/2}\right)}^2-1}{x-x^{1/2}}]}^{10}={[\frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1}-\frac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)}]}^{10}$

$={[x^{1/3}+1-1-x^{1/2}]}^{10}={[x^{1/3}-x^{-1/2}]}^{10}$

General term in the given expansion is

$T_{r+1}={{}^{n}C}_r{\left(x\right)}^r{\left(y\right)}^{n-r}$

${{}^{10}C}_r{\left(x\right)}^{1/3(10-r)}{(-x)}^{-\left(1/2\right)r}={(-1)}^{-r/2}{{}^{10}C}_r{\left(x\right)}^{\frac{10-r}{3}+\left(\frac{-1}{2}r\right)}={(-1)}^{-r/2}{{}^{10}C}_r{\left(x\right)}^{\frac{20-2r-3r}{6}}$

since required term is independent of $x$

$\therefore \frac{20-2r-3r}{6}=0\Rightarrow -5r=-20\Rightarrow r=4$

$\therefore$ coeffcient of term independent of $x$ is

$T_{4+1}=T_5={{}^{10}C}_4{(-1)}^{-4/2}={{}^{10}C}_4{(-1)}^{-2}$

$=\frac{10!}{6!\times 4!}=210$