Question

In the expansion of ${(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}})}^{10}$, the 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

Answer: (C), (D)

The correct options are
C ${}^{10}{C}_4$
D ${}^{10}{C}_6$

${(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}})}^{10}={(\frac{(x+1)(x^{\frac{1}{3}}+1)}{({\left(x^{\frac{1}{3}}\right)}^2-\left(x^{\frac{1}{3}}\right)\times 1+1)(x^{\frac{1}{3}}+1)}-\frac{x-1}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)})}^{10}={((1+x^{\frac{1}{3}})-\frac{(x^{\frac{1}{2}}+1)}{x^{\frac{1}{2}}})}^{10}={(x^{\frac{1}{3}}-x^{-\frac{1}{2}})}^{10}$

Now,

$T_{r+1}={}^{10}{C}_r{x}^{\frac{10-r}{3}}{x}^{\frac{-r}{2}}$

For the term to not contain any power of $x$,

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

$\Rightarrow r=4$

Hence the terms are ${}^{10}{C}_4$ and ${}^{10}{C}_6$