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Question Number 166620 by mathlove last updated on 23/Feb/22

Answered by benhamimed last updated on 23/Feb/22

x+(1/x)≠(√3) ∀ x∈R^∗ (x+(1/x))′=1−(1/x^2 )=((x^2 −1)/x^2 ) (x+(1/x))∈]−∞;−2]∪[2;+∞[

$${x}+\frac{\mathrm{1}}{{x}}\neq\sqrt{\mathrm{3}}\:\:\:\:\:\forall\:{x}\in\mathbb{R}^{\ast} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)'=\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$$\left.\left(\left.{x}+\frac{\mathrm{1}}{{x}}\right)\in\right]−\infty;−\mathrm{2}\right]\cup\left[\mathrm{2};+\infty\left[\right.\right. \\ $$

Answered by Rasheed.Sindhi last updated on 23/Feb/22

x+(1/x)=(√3) ; x^(18) +x^(12) +x^6 +1=? •(x+(1/x))^3 =x^3 +(1/x^3 )+3(x+(1/x)) x^3 +(1/x^3 )=((√3) )^3 −3((√3) )=3^(3/2) −3^(3/2) =0 •(x^3 +(1/x^3 ))^3 =x^9 +(1/x^9 )+3(x^3 +(1/x^3 )) x^9 +(1/x^9 )=(0)^3 −3(0)=0 •x^(18) +x^(12) +x^6 +1=x^9 ((( x^(18) +x^(12) +x^6 +1)/x^9 )) =x^9 (x^9 +(1/x^9 )+x^3 +(1/x^3 ))=x^9 (0+0)=0

$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}\:\:;\:\:\:{x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1}=? \\ $$$$\bullet\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} −\mathrm{3}\left(\sqrt{\mathrm{3}}\:\right)=\mathrm{3}^{\mathrm{3}/\mathrm{2}} −\mathrm{3}^{\mathrm{3}/\mathrm{2}} =\mathrm{0} \\ $$$$\bullet\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} ={x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+\mathrm{3}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right) \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }=\left(\mathrm{0}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$ \\ $$$$\:\bullet{x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1}={x}^{\mathrm{9}} \left(\frac{\:{x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{9}} }\right) \\ $$$$={x}^{\mathrm{9}} \left({x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)={x}^{\mathrm{9}} \left(\mathrm{0}+\mathrm{0}\right)=\mathrm{0} \\ $$

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