The term independent of x(x0, x neq 1) in th | vedclass

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(B) Given expression: $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$ Simplify the terms inside the br

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(B) Given expression: $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1ight)}-\frac{(x-1)}{(x-\sqrt{x})}ight]^{10}$ Simplify the terms inside the bracket: $\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1ight)} = \frac{(x^{1/3})^3 + 1^3}{x^{2/3} - x^{1/3} + 1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3} + 1$ $\frac{(x-1)}{(x-\sqrt{x})} = \frac{(\sqrt{x})^2 - 1}{\sqrt{x}(\sqrt{x}-1)} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$ Substituting these back: $\left[(x^{1/3} + 1) - (1 + x^{-1/2})ight]^{10} = (x^{1/3} - x^{-1/2})^{10}$ The general term $T_{r+1}$ is given by: $T_{r+1} = {}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$ For the term independent of $x$,the exponent of $x$ must be $0$: $\frac{10-r}{3} - \frac{r}{2} = 0$ $\Rightarrow 20 - 2r - 3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$ The term is $T_{4+1} = {}^{10}C_4 (-1)^4 = {}^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$

The term independent of $x(x>0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$ is:

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