Question

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

The correct option is D ${}^{10}{C}_6$

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

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

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

$[a^3+b^3=(a+b)(a^3+b^2-ab){\left(x^{1/3}\right)}^3+1^3=(x^{1/3}+1)(x^{2/3}+1-x^{1/3})x+1=]$

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

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

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

For term without $x$

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

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

$r=6$

Thus $T_7{=}^{10}{C}_6{x}^0$

$T_7{=}^{10}{C}_6\Rightarrow$ term without $x$