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Question Number 9747 by tawakalitu last updated on 30/Dec/16

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

Answered by sandy_suhendra last updated on 30/Dec/16

$$\mathrm{using}\mathrm{the}\mathrm{formula}:$$$${\mathrm{a}}^3+{\mathrm{b}}^3=(\mathrm{a}+\mathrm{b})({\mathrm{a}}^2-\mathrm{ab}+{\mathrm{b}}^2)$$$${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$$$$$$$${[\frac{{\left({\mathrm{x}}^{1/3}\right)}^3+1^3}{{\mathrm{x}}^{2/3}-{\mathrm{x}}^{1/3}+1}-\frac{{\left({\mathrm{x}}^{1/2}\right)}^2-1^2}{{\mathrm{x}}^{1/2}({\mathrm{x}}^{1/2}-1)}]}^{10}$$$$={[\frac{({\mathrm{x}}^{1/3}+1)({\mathrm{x}}^{2/3}-{\mathrm{x}}^{1/3}+1)}{{\mathrm{x}}^{2/3}-{\mathrm{x}}^{1/3}+1}-\frac{({\mathrm{x}}^{1/2}+1)({\mathrm{x}}^{1/2}-1)}{{\mathrm{x}}^{1/2}({\mathrm{x}}^{1/2}-1)}]}^{10}$$$$={[{\mathrm{x}}^{1/3}+1-1-\frac{1}{{\mathrm{x}}^{1/2}}]}^{10}$$$$={[{\mathrm{x}}^{1/3}-\frac{1}{{\mathrm{x}}^{1/2}}]}^{10}$$$$\mathrm{the}\mathrm{term}\mathrm{independent}\mathrm{of}\mathrm{x}={\mathrm{x}}^0$$$$\Rightarrow 10\mathrm{C}6{\left({\mathrm{x}}^{1/3}\right)}^6{(-\frac{1}{{\mathrm{x}}^{1/2}})}^4=210{\mathrm{x}}^0$$$$\mathrm{so}\mathrm{the}\mathrm{coefficient}\mathrm{is}210$$$$$$

Commented by tawakalitu last updated on 30/Dec/16

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

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