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Question Number 115399 by EvoneAkashi last updated on 25/Sep/20

$$\mathrm{The}\mathrm{coefficient}\mathrm{of}\mathrm{the}\mathrm{term}\mathrm{independent}$$$$\mathrm{of}x\mathrm{in}\mathrm{the}\mathrm{expansion}\mathrm{of}$$$${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}})}^{10}\mathrm{is}$$

Answered by Dwaipayan Shikari last updated on 25/Sep/20

$${(\frac{\mathrm{x}+1}{(\mathrm{x}+1)}.({\mathrm{x}}^{\frac{1}{3}}+1)-\frac{({\mathrm{x}}^{\frac{1}{2}}+1)({\mathrm{x}}^{\frac{1}{2}}-1)}{{\mathrm{x}}^{\frac{1}{2}}({\mathrm{x}}^{\frac{1}{2}}-1)})}^{10}$$$$={({\mathrm{x}}^{\frac{1}{3}}+1-1-{\mathrm{x}}^{-\frac{1}{2}})}^{10}$$$$={({\mathrm{x}}^{\frac{1}{3}}-{\mathrm{x}}^{-\frac{1}{2}})}^{10}={\mathrm{x}}^{-5}{({\mathrm{x}}^{\frac{5}{6}}-1)}^{10}$$$${\mathrm{T}}_{\mathrm{r}+1}={\mathrm{x}}^{-5}\left(\begin{array}{c}10\\ \mathrm{r}\end{array}\right){\mathrm{x}}^{\frac{5\mathrm{r}}{6}}{(-1)}^{10-\mathrm{r}}$$$${\left(\mathrm{x}\right)}^{\frac{5\mathrm{r}}{6}}={\mathrm{x}}^5$$$$\mathrm{r}=6$$$${\mathrm{T}}_{6+1}\mathrm{Coefficient}=\left(\frac{10!}{6!4!}\right)=210$$

Commented by mr W last updated on 25/Sep/20

$$whatifthequestionis$$$${(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}+1})}^{10}$$

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