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Q.
The coefficient of the term independent of $x$ in the expansion of $\left(\frac{ x +1}{ x ^{2 / 3}- x ^{1 / 3}+1}-\frac{ x -1}{ x - x ^{1 / 2}}\right)$ is
Binomial Theorem
Solution:
We have,
$\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}$
$=\frac{\left(x^{1 / 3}\right)^{3}+1^{3}}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x^{1 / 2}\left(x^{1 / 2}-1\right)}$
$=\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x^{1 / 2}+1}{x^{1 / 2}}$
$=x^{1 / 3}+1-1-x^{-1 / 2}=x^{1 / 3}-x^{-1 / 2}$
$\left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$
Let $T_{r+1}$ be the general term in $\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}$.
Then, $T _{ r +1}={ }^{10} C _{ r }\left( x ^{1 / 3}\right)^{10- r }(-1)^{ r }\left( x ^{-1 / 2}\right)^{ r }$
For this term to be independent of $x$, we must have $\frac{10- r }{3}-\frac{ r }{2}=0$
or $20-2 r -3 r =0$
or $r =4$
So, the required coefficient is ${ }^{10} C _{4}(-1)^{4}=210$.