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Question Number 60905 by necx1 last updated on 27/May/19

if x+(1/x)=(√3).find x^(24) +x^(18) +x^6 +1

$${if}\:{x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}.{find} \\ $$$${x}^{\mathrm{24}} +{x}^{\mathrm{18}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$

Answered by ajfour last updated on 27/May/19

E=x^(24) +x^(18) +x^6 +1 x^2 −(√3)x+1=0 x=(((√3)±i)/2) = e^(±i (π/6)) E= 1−1−1+1=0 .

$${E}={x}^{\mathrm{24}} +{x}^{\mathrm{18}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$\:\:\:{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{x}=\frac{\sqrt{\mathrm{3}}\pm{i}}{\mathrm{2}}\:=\:{e}^{\pm{i}\:\frac{\pi}{\mathrm{6}}} \\ $$$${E}=\:\mathrm{1}−\mathrm{1}−\mathrm{1}+\mathrm{1}=\mathrm{0}\:. \\ $$

Commented by necx1 last updated on 27/May/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 27/May/19

x+(1/x)=(√3) (Given) (x+(1/x))^2 =((√3) )^2 x^2 +(1/x^2 )=3−2=1 (x^2 +(1/x^2 ))^3 =(1)^3 =1 (x^2 )^3 +(1/((x^2 )^3 ))+3.x^2 .(1/x^2 )(x^2 +(1/x^2 ))=1 x^6 +(1/x^6 )=1−3=−2........................A (x^6 +(1/x^6 ))^2 =(−2)^2 =4 x^(12) +(1/x^(12) )=4−2=2......................B ⇒x^(24) −2x^(12) +1=0 ⇒(x^(12) −1)^2 =0⇒x^(12) =1...........C Now, Let P=x^(24) +x^(18) +x^6 +1 P=x^(12) (x^(12) +(1/x^(12) )+x^6 +(1/x^6 )) =(1)(2−2)=0

$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}\:\:\:\:\left({Given}\right) \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\:\:\right)^{\mathrm{2}} \:\: \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{3}−\mathrm{2}=\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{3}} =\left(\mathrm{1}\right)^{\mathrm{3}} =\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} }+\mathrm{3}.{x}^{\mathrm{2}} .\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }=\mathrm{1}−\mathrm{3}=−\mathrm{2}........................{A} \\ $$$$\left({x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)^{\mathrm{2}} =\left(−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{12}} +\frac{\mathrm{1}}{{x}^{\mathrm{12}} }=\mathrm{4}−\mathrm{2}=\mathrm{2}......................{B} \\ $$$$\:\Rightarrow{x}^{\mathrm{24}} −\mathrm{2}{x}^{\mathrm{12}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{12}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{x}^{\mathrm{12}} =\mathrm{1}...........{C} \\ $$$${Now}, \\ $$$${Let}\:{P}={x}^{\mathrm{24}} +{x}^{\mathrm{18}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$${P}={x}^{\mathrm{12}} \left({x}^{\mathrm{12}} +\frac{\mathrm{1}}{{x}^{\mathrm{12}} }+{x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right) \\ $$$$\:\:\:\:=\left(\mathrm{1}\right)\left(\mathrm{2}−\mathrm{2}\right)=\mathrm{0} \\ $$

Commented by necx1 last updated on 27/May/19

too many methods for approaching the problem.I′m so grateful.

$${too}\:{many}\:{methods}\:{for}\:{approaching}\:{the} \\ $$$${problem}.{I}'{m}\:{so}\:{grateful}. \\ $$

Answered by math1967 last updated on 27/May/19

x+(1/x)=(√3) ⇒x^3 +(1/x^3 )=0 ((x^6 +1)/x^3 )=0 ∴x^6 +1=0 ∴x^(24) +x^(18) +x^6 +1=x^(18) (x^6 +1)+x^6 +1 =x^(18) ×0+0=0ans

$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}\:\Rightarrow{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{0}\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\therefore{x}^{\mathrm{6}} +\mathrm{1}=\mathrm{0}\:\:\therefore{x}^{\mathrm{24}} +{x}^{\mathrm{18}} +{x}^{\mathrm{6}} +\mathrm{1}={x}^{\mathrm{18}} \left({x}^{\mathrm{6}} +\mathrm{1}\right)+{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}^{\mathrm{18}} ×\mathrm{0}+\mathrm{0}=\mathrm{0}{ans} \\ $$

Commented by Rasheed.Sindhi last updated on 27/May/19

More efficient approach than mine!

$$\mathcal{M}{ore}\:{efficient}\:{approach}\:{than}\:{mine}! \\ $$

Commented by math1967 last updated on 27/May/19

Thank you sir

$${Thank}\:{you}\:{sir} \\ $$

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