The term independent of x in the expansion of | vedclass

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Question

(A) Let the expression be $E = \left(\frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}}\right)^{10}$.Using the formula $a^3+b^3 = (a+b)(a^2-ab+b^2)

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(A) Let the expression be $E = \left(\frac{x+1}{x^{2/3}-x^{1/3}+1} - \frac{x-1}{x-x^{1/2}}ight)^{10}$.Using the formula $a^3+b^3 = (a+b)(a^2-ab+b^2)$,we have $\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$.For 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,$E = \left((x^{1/3}+1) - (1+x^{-1/2})ight)^{10} = \left(x^{1/3}-x^{-1/2}ight)^{10}$.The general term $T_{r+1}$ is given by ${}^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = {}^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$.For the term to be independent of $x$,the exponent must be zero: $\frac{10-r}{3} - \frac{r}{2} = 0$.$2(10-r) - 3r = 0 \implies 20 - 5r = 0 \implies r = 4$.The term is ${}^{10}C_4 (-1)^4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$.

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

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