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Question-195135

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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 +(1/x^3 )=(x+(1/x))^3 −3x.(1/x)(x+(1/x)) =3(√3)−3(√3)=0 ((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) ×0=0
$$\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}{x}.\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\:{x}^{\mathrm{6}} +\mathrm{1}=\mathrm{0} \\ $$$$\:{x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$={x}^{\mathrm{12}} \left({x}^{\mathrm{6}} +\mathrm{1}\right)+\mathrm{0} \\ $$$$={x}^{\mathrm{12}} ×\mathrm{0}=\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jul/23
x+(1/x)=(√3) , x^(18) +x^(12) +x^6 +1=? (x+(1/x))^3 =((√3) )^3 =3(√3) x^3 +(1/x^3 )+3(x+(1/x))=3(√3) x^3 +(1/x^3 )=3(√3) −3((√3) )=0 (x^3 +(1/x^3 ))^3 =(0)^3 x^9 +(1/x^9 )+3(x^3 +(1/x^3 ))=0 x^9 +(1/x^9 )=0 x^(18) +x^(12) +x^6 +1 =x^9 (x^9 +(1/x^9 )+x^3 +(1/x^3 )) =x^9 (0+0)=0
$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}\:\:,\:{x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1}=? \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} =\mathrm{3}\sqrt{\mathrm{3}}\: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\mathrm{3}\sqrt{\mathrm{3}}\: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{3}\left(\sqrt{\mathrm{3}}\:\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{3}} =\left(\mathrm{0}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+\mathrm{3}\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\mathrm{0} \\ $$$${x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }=\mathrm{0} \\ $$$${x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$={x}^{\mathrm{9}} \left({x}^{\mathrm{9}} +\frac{\mathrm{1}}{{x}^{\mathrm{9}} }+{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right) \\ $$$$={x}^{\mathrm{9}} \left(\mathrm{0}+\mathrm{0}\right)=\mathrm{0} \\ $$
Answered by BaliramKumar last updated on 25/Jul/23
x + (1/x) = (√3) x + (1/x) = 2(((√3)/2)) x + (1/x) = 2cos((π/6)) x^n + (1/x^n ) = 2cos(((nπ)/6)) x^3 + (1/x^3 ) = 2cos(((3π)/6)) = 0 x^9 + (1/x^9 ) = 2cos(((9π)/6)) = 0 x^(18) + x^(12) + x^6 + 1 ⇒ x^(18) + 1 + x^(12) + x^6 x^9 (x^9 + (1/x^9 )) + x^9 (x^3 + (1/x^3 )) x^9 (0) + x^9 (0) = 0
$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\sqrt{\mathrm{3}} \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${x}\:+\:\frac{\mathrm{1}}{{x}}\:=\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${x}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{n}} }\:=\:\mathrm{2}{cos}\left(\frac{{n}\pi}{\mathrm{6}}\right) \\ $$$${x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{3}\pi}{\mathrm{6}}\right)\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{9}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\:=\:\mathrm{2}{cos}\left(\frac{\mathrm{9}\pi}{\mathrm{6}}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}^{\mathrm{18}} \:+\:{x}^{\mathrm{12}} \:+\:{x}^{\mathrm{6}} \:+\:\mathrm{1}\:\Rightarrow\:{x}^{\mathrm{18}} \:+\:\mathrm{1}\:+\:{x}^{\mathrm{12}} \:+\:{x}^{\mathrm{6}} \\ $$$${x}^{\mathrm{9}} \left({x}^{\mathrm{9}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{9}} }\right)\:+\:{x}^{\mathrm{9}} \left({x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right) \\ $$$${x}^{\mathrm{9}} \left(\mathrm{0}\right)\:+\:{x}^{\mathrm{9}} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 25/Jul/23
A unique approach!
$$\mathrm{A}\:\mathrm{unique}\:\mathrm{approach}! \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jul/23
x+(1/x)=(√3) ⇒x^2 =(√3) x−1 ⇒x^3 =(√3) x^2 −x=(√3) ((√3) x−1)−x =2x−(√3) ⇒x^6 =(2x−(√3))^2 =4x^2 −4(√3) x+3 =4((√3) x−1)−4(√3) x+3 =−1 x^(18) +x^(12) +x^6 +1 =(x^6 )^3 +(x^6 )^2 +x^6 +1 =(−1)^3 +(−1)^2 +(−1)+1=0
$${x}+\frac{\mathrm{1}}{{x}}=\sqrt{\mathrm{3}}\:\Rightarrow{x}^{\mathrm{2}} =\sqrt{\mathrm{3}}\:{x}−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} =\sqrt{\mathrm{3}}\:{x}^{\mathrm{2}} −{x}=\sqrt{\mathrm{3}}\:\left(\sqrt{\mathrm{3}}\:{x}−\mathrm{1}\right)−{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{x}−\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}^{\mathrm{6}} =\left(\mathrm{2}{x}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\:{x}+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\left(\sqrt{\mathrm{3}}\:{x}−\mathrm{1}\right)−\mathrm{4}\sqrt{\mathrm{3}}\:{x}+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{1} \\ $$$${x}^{\mathrm{18}} +{x}^{\mathrm{12}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$=\left({x}^{\mathrm{6}} \right)^{\mathrm{3}} +\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} +{x}^{\mathrm{6}} +\mathrm{1} \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{3}} +\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$

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