JEE Main · 2013 · 07 Apr 2013 · Offline · Binomial Theorem

Simplify each fraction first. Using $a^3+1=(a+1)(a^2-a+1)$ with $a=x^{1/3}$, the first term $\frac{x+1}{x^{2/3}-x^{1/3}+1}=x^{1/3}+1$. The second, $\frac{x-1}{x-x^{1/2}}=\frac{(x^{1/2}+1)(x^{1/2}-1)}{x^{1/2}(x^{1/2}-1)}=1+x^{-1/2}$. Their difference is $x^{1/3}-x^{-1/2}$, so the bracket is ${\left(x^{1/3}-x^{-1/2}\right)}^{10}$. The general term $\left(\genfrac{}{}{0ex}{}{10}{r}\right)(-1{)}^r{x}^{\frac{20-5r}{6}}$ is free of $x$ when $20-5r=0$, i.e. $r=4$, giving the coefficient $\left(\genfrac{}{}{0ex}{}{10}{4}\right)=210$.