Question

In the expansions of ${(\frac{x+1}{x^{2/3}+x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$, then term which does not contain $x$, is equal to

• A. ${}^{10}{C}_0$
• B. ${}^{10}{C}_7$
• C. ${}^{10}{C}_4$
• D. ${}^{10}{C}_6$

Answer 1

The correct option is C ${}^{10}{C}_4$

${(\frac{x-1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}$

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

${(x^{1/3}-x^{-1/2})}^{10}$

Greneral term

${}^{10}{C}_r{\left(x^{\frac{1}{3}}\right)}^{10-r}\cdot (-1{)}^r\left(x^{-\frac{r}{2}}\right)$

Power of $x=0$

$\frac{10-r}{3}-\frac{r}{2}=0$

$20-2r-3r=0$

$20=5r$

$r=4$

$\therefore T_{4+1}={}^{10}{C}_4(-1{)}^4$

$={}^{10}{C}_4=5^{\text{th}}\text{term}$

option (c)