Question

The term independent of $x(x>0,x$ $\ne$ $1)$ in the expansion of ${\left[\frac{(x+1)}{(x^{2/3}-x^{1/3}+1)}-\frac{(x-1)}{(x-\sqrt{x})}\right]}^{10}$ is

• A. 105
• B. 210
• C. 315
• D. 420

Answer 1

Answer: (B)

210

$\frac{(x+1)}{(x^2/3-x^{1/3}+1)}-\frac{(x-1)}{(x-\sqrt{x})}$

$(x+1)=(x^{1/3}+1^{1/3})(x^{2/3}-x^{1/3}+1)$ ...(1)

$(x-\sqrt{x})=\sqrt{x}(\sqrt{x}-1)$ ....(2)

$(x-1)=\left[\right(\sqrt{x}{)}^2-1]$

$(\sqrt{x}+1)(\sqrt{x}-1)$ ...(3)

$(x^{1/3}+1^{1/3})-\frac{(\sqrt{x}+1)}{\sqrt{x}}$

$(x^{1/3}+1)-x^{-1/2}[x^{1/2}+1]$

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

$[x^{1/3}+1-1+(x{)}^{-1/2}]$

$T_{r+1}{=}^n{C}_r\left(x{)}^{r/3}{x}^{-1/2}\right(n-r)$

$\frac{r}{3}-\frac{10+r}{2}$

So, $\frac{r}{3}-5+\frac{r}{2}=0$

$\Rightarrow \frac{5}{6}r=5\Rightarrow r=6$

So, the term is ${}^n{C}_r{=}^{10}{C}_6$

${\Rightarrow }^{10}{C}_6=\frac{10!}{6!4!}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}$

$=210$