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The term independent of $x$ in the expansion $\bigg(\large\frac{x+1}{x^{\large\frac{2}{3}}-x^{\large\frac{1}{3}}+1}-\frac{x-1}{x-\sqrt x}\bigg)^{10}$ is ?

$\begin{array}{1 1} 4 \\ 120 \\ 210 \\ 0 \end{array} $

1 Answer

$x+1=(x^{\frac{1}{3}})^3+1^3$$=(x^{\frac{1}{3}}+1)$$(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1)$
$\therefore \:\large\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}$$=x^{\frac{1}{3}}+1$
similarly $\large\frac{x-1}{x-\sqrt x}=\frac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x(\sqrt x-1)}$$=\large\frac{\sqrt x+1}{\sqrt x}$$=1+x^{1/2}$
Now
$\bigg(\large\frac{x+1}{x^{2\3}-x^{1\3}+1}-\large\frac{x-1}{x-\sqrt x}\bigg)^{10}=\big(x^{1/3}+1-1-x^{-1/2}\big)^{10}$
$=\big(x^{1/3}-x^{-1/2}\big)^{10}$
General term $T_{r+1}=(-1)^r.^{10}C_r.x^{(10-r)/3}.x^{-r/2}$
$=(-1)^r.^{10}C_r.x^{(20-5r)/6}$
For independent term $\large\frac{20-5r}{6}=0$
$\Rightarrow\:r=4$
$\therefore$ The independent term $=^{10}C_4=210$
answered Dec 1, 2013 by rvidyagovindarajan_1
 

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