The term independent of $x(x0, x
eq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1 ight)}-\frac{(x-1)}{(x-\sqrt{x})} ight]^{10}$

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & {\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1ight)}-\frac{(x-1)}{(x-\sqrt{x})}ight]^{10}} \ & \quad=\left[\frac{\left(x^{1 / 3}ight)^3+1^3}{\left(x^{2 / 3}-x^{1 / 3}+1ight)}-\frac{\left\{(\sqrt{x})^2-1ight\}}{\sqrt{x}(\sqrt{x}-1)}ight]^{10} \ & \quad=\left[\frac{\left(x^{1 / 3}+1ight)\left(x^{2 / 3}-x^{1 / 3}+1ight)}{\left(x^{2 / 3}-x^{1 / 3}+1ight)}-\frac{\left\{(\sqrt{x})^2-1ight\}}{\sqrt{x}(\sqrt{x}-1)}ight]^{10} \ & \quad=\left[\left(x^{1 / 3}+1ight)-\frac{(\sqrt{x}+1)}{\sqrt{x}}ight]^{10}=\left(x^{1 / 3}-x^{-1 / 2}ight)^{10}\end{aligned}$ The general term is $ \begin{aligned} T_{r+1} & { }^{10} C_r\left(x^{1 / 3}ight)^{10-r}\left(-x^{-1 / 2}ight)^r \ & ={ }^{10} C_r(-1)^r \cdot x^{\frac{10-r}{3}-\frac{r}{2}} \end{aligned} $ For the term independent of $x$, put $ \begin{array}{cc} & \frac{10-r}{3}-\frac{r}{2}=0 \ \Rightarrow & 20-2 r-3 r=0 \ \Rightarrow & 20=5 r \Rightarrow r=4 \ 	herefore & T_5={ }^{10} C_4=\frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1}=210 \end{array} $

The term independent of $x(x>0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 /…

Solution