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

Answered by benhamimed last updated on 23/Feb/22

$$x+\frac{1}{x}\ne \sqrt{3}\mathrm{\forall }x\in {\mathbb{R}}^{\ast }$$$${(x+\frac{1}{x})}^{\prime }=1-\frac{1}{x^2}=\frac{x^2-1}{x^2}$$$$(x+\frac{1}{x})\in ]-\mathrm{\infty };-2]\cup [2;+\mathrm{\infty }[$$

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

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

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