Question-173042

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

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

(x+(1/x))^2 =x^2 +(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⇒(x^2 )^3 +1=0 determinant (((x^6 +1=0))) 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) =(0)(x^(200) +x^(84) +x^(12) +1) =0

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

Commented by mathlove last updated on 06/Jul/22

$${thanks}\:{sir} \\ $$

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

(x+(1/x))^2 =3 x^2 +2+(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 =(x^6 )^(34) x^2 +(x^6 )^(33) x^2 +(x^6 )^(15) +(x^6 )^(14) +(x^6 )^3 +(x^6 )^2 +(−1)+1 =(−1)^(34) x^2 +(−1)^(33) x^2 +(−1)^(15) +(−1)^(14) +(−1)^3 +(−1)^2 +(−1)+1 =x^2 −x^2 −1+1−1+1−1+1=0

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

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