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Question Number 173042 by mathlove last updated on 05/Jul/22

Answered by Rasheed.Sindhi last updated on 05/Jul/22

$${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2=3$$$$x^4-x^2+1=0$$$$x^2=y:y^2-y+1=0$$$$(y+1)(y^2-y+1)=0$$$$y^3+1=0\Rightarrow {\left(x^2\right)}^3+1=0$$$$\overline{)\begin{array}{c}{x}^6+1=0\end{array}}$$$$$$$$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$$$$=x^{200}(x^6+1)+x^{84}(x^6+1)+x^{12}(x^6+1)+(x^6+1)$$$$=(x^6+1)(x^{200}+x^{84}+x^{12}+1)$$$$=\left(0\right)(x^{200}+x^{84}+x^{12}+1)$$$$=0$$

Commented by mathlove last updated on 06/Jul/22

$$thankssir$$

Answered by Rasheed.Sindhi last updated on 05/Jul/22

$${(x+\frac{1}{x})}^2=3$$$$x^2+2+\frac{1}{x^2}=3$$$$x^4-x^2+1=0$$$$x^4=x^2-1$$$$x^6=x^4-x^2=x^2-1-x^2=-1$$$$$$$$x^{206}+x^{200}+x^{90}+x^{84}+x^{18}+x^{12}+x^6+1$$$$={\left(x^6\right)}^{34}{x}^2+{\left(x^6\right)}^{33}{x}^2+{\left(x^6\right)}^{15}+{\left(x^6\right)}^{14}+{\left(x^6\right)}^3+{\left(x^6\right)}^2+(-1)+1$$$$={(-1)}^{34}{x}^2+{(-1)}^{33}{x}^2+{(-1)}^{15}+{(-1)}^{14}+{(-1)}^3+{(-1)}^2+(-1)+1$$$$=\cancel{x^2}-\cancel{x^2}-\cancel{1}+\cancel{1}-\cancel{1}+\cancel{1}-\cancel{1}+\cancel{1}=0$$

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