The term independent of 'x' in the expansion | vedclass

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Question

(B) Simplify the expression inside the bracket: $\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}

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(B) Simplify the expression inside the bracket: $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$ $\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1 + x^{-1/2}$ So,the expression becomes: $(x^{1/3}+1 - (1 + x^{-1/2}))^{10} = (x^{1/3} - x^{-1/2})^{10}$ 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 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 independent term is: $T_{4+1} = {}^{10}C_4 (-1)^4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$

The term independent of '$x$' in the expansion of $\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)^{10}$,where $x \neq 0, 1$,is equal to $.....$

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