In the expansion of left( frac{x + 1}{x^{2/3} | vedclass

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(C) Let the given expression be $E = \left( \frac{x + 1}{x^{2/3} - x^{1/3} + 1} - \frac{x - 1}{x - x^{1/2}} \right)^{10}$.First,simplify the first ter

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$^{10}C_4$

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(C) Let the given expression be $E = \left( \frac{x + 1}{x^{2/3} - x^{1/3} + 1} - \frac{x - 1}{x - x^{1/2}} \right)^{10}$.First,simplify the first term: $\frac{x + 1}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3} + 1)(x^{2/3} - x^{1/3} + 1)}{x^{2/3} - x^{1/3} + 1} = x^{1/3} + 1$.Next,simplify the second term: $\frac{x - 1}{x - x^{1/2}} = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x}(\sqrt{x} - 1)} = \frac{\sqrt{x} + 1}{\sqrt{x}} = 1 + x^{-1/2}$.Substituting these back into the expression: $E = (x^{1/3} + 1 - (1 + x^{-1/2}))^{10} = (x^{1/3} - x^{-1/2})^{10}$.The general term $T_{r+1}$ in the expansion of $(x^{1/3} - x^{-1/2})^{10}$ is given by $T_{r+1} = ^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = ^{10}C_r (-1)^r x^{(10-r)/3} x^{-r/2} = ^{10}C_r (-1)^r x^{(20-2r-3r)/6} = ^{10}C_r (-1)^r x^{(20-5r)/6}$.For the term to be independent of $x$,the exponent must be zero: $\frac{20-5r}{6} = 0$ $\Rightarrow 20 - 5r = 0$ $\Rightarrow r = 4$.The term is $T_{4+1} = T_5 = ^{10}C_4 (-1)^4 = ^{10}C_4$.

In the expansion of $\left( \frac{x + 1}{x^{2/3} - x^{1/3} + 1} - \frac{x - 1}{x - x^{1/2}} \right)^{10}$,the term which does not contain $x$ is:

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