Mathematics - Binomial Theorem Question with Solution | TestHub

Given expression $=\frac{{\left(x^{1/3}\right)}^3+(1{)}^3}{x^{2/3}-x^{1/3}+1}-\frac{(x-1)}{x^{1/2}\left(x^{1/2}-1\right)}$ $=\frac{\left(x^{1/3}+1\right)\left(x^{2/3}-x^{1/3}+1\right)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{\left(x^{1/2}+1\right)\left(x^{1/2}-1\right)}{x^{1/2}\left(x^{1/2}-1\right)}$ $=\left(x^{1/3}+1\right)-\left(1+x^{-1/2}\right)=x^{1/3}-x^{-1/2}$ $\Rightarrow {\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)}^{10}$ $={\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)}^{10}$ $={\left(x^{1/3}-x^{-1/2}\right)}^{10}$ $T_{r+1}$ in ${\left(x^{1/3}-x^{-1/2}\right)}^{10}$ is ${}^{10}{C}_r{\left(x^{1/3}\right)}^{10-r}\cdot (-1{)}^r\cdot {\left(x^{-1/2}\right)}^r$ $=(-1{)}^r{}^{10}{C}_r{x}^{\left(\frac{10-r}{3}-\frac{r}{2}\right)}$ which is independent of x If $\left(\frac{10-r}{3}-\frac{r}{2}\right)=0\Rightarrow r=4$ Hence required coefficient $={}^{10}{C}_4(-1{)}^4=210$