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Question Number 195135 by 073 last updated on 25/Jul/23

Answered by som(math1967) last updated on 25/Jul/23

$$x^3+\frac{1}{x^3}={(x+\frac{1}{x})}^3-3x.\frac{1}{x}(x+\frac{1}{x})$$$$=3\sqrt{3}-3\sqrt{3}=0$$$$\frac{x^6+1}{x^3}=0$$$$x^6+1=0$$$$x^{18}+x^{12}+x^6+1$$$$=x^{12}(x^6+1)+0$$$$=x^{12}\times 0=0$$

Answered by Rasheed.Sindhi last updated on 25/Jul/23

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

Answered by BaliramKumar last updated on 25/Jul/23

$$x+\frac{1}{x}=\sqrt{3}$$$$x+\frac{1}{x}=2\left(\frac{\sqrt{3}}{2}\right)$$$$x+\frac{1}{x}=2cos\left(\frac{\pi }{6}\right)$$$$x^n+\frac{1}{x^n}=2cos\left(\frac{n\pi }{6}\right)$$$$x^3+\frac{1}{x^3}=2cos\left(\frac{3\pi }{6}\right)=0$$$$x^9+\frac{1}{x^9}=2cos\left(\frac{9\pi }{6}\right)=0$$$$$$$$x^{18}+x^{12}+x^6+1\Rightarrow x^{18}+1+x^{12}+x^6$$$$x^9(x^9+\frac{1}{x^9})+x^9(x^3+\frac{1}{x^3})$$$$x^9\left(0\right)+x^9\left(0\right)=0$$$$$$$$$$

Commented by Rasheed.Sindhi last updated on 25/Jul/23

$$\mathrm{A}\mathrm{unique}\mathrm{approach}!$$

Answered by Rasheed.Sindhi last updated on 25/Jul/23

$$x+\frac{1}{x}=\sqrt{3}\Rightarrow x^2=\sqrt{3}x-1$$$$\Rightarrow x^3=\sqrt{3}{x}^2-x=\sqrt{3}(\sqrt{3}x-1)-x$$$$=2x-\sqrt{3}$$$$\Rightarrow x^6={(2x-\sqrt{3})}^2=4x^2-4\sqrt{3}x+3$$$$=4(\sqrt{3}x-1)-4\sqrt{3}x+3$$$$=-1$$$$x^{18}+x^{12}+x^6+1$$$$={\left(x^6\right)}^3+{\left(x^6\right)}^2+x^6+1$$$$={(-1)}^3+{(-1)}^2+(-1)+1=0$$

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