The term independent of x in the expansion of | vedclass

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(C) Simplify the expression inside the bracket: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/

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(C) Simplify the expression inside the bracket:First term: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$Second term: $\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$Subtracting the terms: $(x^{1/3}+1) - (1 + x^{-1/2}) = x^{1/3} - x^{-1/2}$The expression becomes $(x^{1/3} - x^{-1/2})^{10}$.The general term $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.For the term to be independent of $x$,the exponent must be zero:$\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$.The term is ${}^{10}C_4 (-1)^4 = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.

The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}, x \neq 1,$ is equal to ....... .

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