Question Number 199731 by Rupesh123 last updated on 08/Nov/23
Answered by des_ last updated on 08/Nov/23
$$\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)^{\mathrm{3}} =\:\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\mathrm{3}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{3}\sqrt{\mathrm{3}}\:\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{0}\:\:\Rightarrow\:{x}^{\mathrm{6}} \:=\:−\mathrm{1}; \\ $$$${x}^{\mathrm{2023}} \:=\:\left({x}^{\mathrm{6}} \right)^{\mathrm{337}} {x}\:=\:−{x}; \\ $$$${x}^{\mathrm{2023}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2023}} }\:=\:−\left({x}\:+\:\frac{\mathrm{1}}{{x}}\right)\:=\:−\sqrt{\mathrm{3}} \\ $$
Commented by Rupesh123 last updated on 08/Nov/23
Perfect, sir!
Answered by mathfreak01 last updated on 08/Nov/23
![OR x = cos θ + isin θ x + (1/x) = 2cos θ = (√3) θ = cos^(−1) ((√3)/2) = 30° x^(2023) + (1/x^(2023) ) = 2cos 2023θ = 2cos 2023(30°) = 2cos [(12(168) + 7)×30°] = 2cos [360(168) + (30×70)] = 2cos 210 = −2cos (210 − 180) = −2cos 30° = −(√3)](https://www.tinkutara.com/question/Q199761.png)
$${OR} \\ $$$${x}\:=\:\mathrm{cos}\:\theta\:\:+\:{i}\mathrm{sin}\:\theta\: \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2cos}\:\theta\:=\:\sqrt{\mathrm{3}} \\ $$$$\theta\:=\:\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:\mathrm{30}° \\ $$$$ \\ $$$${x}^{\mathrm{2023}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2023}} }\:=\:\mathrm{2cos}\:\mathrm{2023}\theta \\ $$$$=\:\mathrm{2cos}\:\mathrm{2023}\left(\mathrm{30}°\right)\:=\:\mathrm{2cos}\:\left[\left(\mathrm{12}\left(\mathrm{168}\right)\:+\:\mathrm{7}\right)×\mathrm{30}°\right] \\ $$$$=\:\mathrm{2cos}\:\left[\mathrm{360}\left(\mathrm{168}\right)\:+\:\left(\mathrm{30}×\mathrm{70}\right)\right] \\ $$$$=\:\mathrm{2cos}\:\mathrm{210}\:=\:−\mathrm{2cos}\:\left(\mathrm{210}\:−\:\mathrm{180}\right) \\ $$$$=\:−\mathrm{2cos}\:\mathrm{30}°\:=\:−\sqrt{\mathrm{3}} \\ $$