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Question Number 158883 by akolade last updated on 10/Nov/21
Commented by Rasheed.Sindhi last updated on 10/Nov/21
((√3) +x_(a) )^(12) +((√3) −x_(b) )^(12) =172928;x=? a+b=2(√3) ;ab=3−x^2 =y (say) •a^3 +b^3 =(a+b)^3 −3ab(a+b) =(2(√3) )^3 −3y(2(√3) ) =2(√3) (12−3y)=6(√3) (4−y) •a^6 +b^6 =(a^3 +b^3 )^2 −2(ab)^3 =(6(√3) (4−y))^2 −2y^3 =108(16−8y+y^2 )−2y^3 =1728−864y+108y^2 −2y^3 •a^(12) +b^(12) =(a^6 +b^6 )^2 −2(ab)^6 =(1728−864y+108y^2 −2y^3 )^2 −2y^6 •(1728−864y+108y^2 −2y^3 )^2 −2y^6 =172928 With the help of advanced calculator y=2⇒3−x^2 =2⇒x=±1
$$\left(\underset{{a}} {\underbrace{\sqrt{\mathrm{3}}\:+{x}}}\right)^{\mathrm{12}} +\left(\underset{{b}} {\underbrace{\sqrt{\mathrm{3}}\:−{x}}}\right)^{\mathrm{12}} =\mathrm{172928};{x}=? \\ $$$${a}+{b}=\mathrm{2}\sqrt{\mathrm{3}}\:;{ab}=\mathrm{3}−{x}^{\mathrm{2}} ={y}\:\left({say}\right) \\ $$$$\bullet{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} −\mathrm{3}{y}\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right) \\ $$$$\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{12}−\mathrm{3}{y}\right)=\mathrm{6}\sqrt{\mathrm{3}}\:\left(\mathrm{4}−{y}\right) \\ $$$$\bullet{a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{6}\sqrt{\mathrm{3}}\:\left(\mathrm{4}−{y}\right)\right)^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{108}\left(\mathrm{16}−\mathrm{8}{y}+{y}^{\mathrm{2}} \right)−\mathrm{2}{y}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:=\mathrm{1728}−\mathrm{864}{y}+\mathrm{108}{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}} \\ $$$$\bullet{a}^{\mathrm{12}} +{b}^{\mathrm{12}} =\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{6}} \\ $$$$\:\:=\left(\mathrm{1728}−\mathrm{864}{y}+\mathrm{108}{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{6}} \\ $$$$\bullet\left(\mathrm{1728}−\mathrm{864}{y}+\mathrm{108}{y}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{172928} \\ $$$${With}\:{the}\:{help}\:{of}\:{advanced}\:{calculator} \\ $$$$\:\:\:{y}=\mathrm{2}\Rightarrow\mathrm{3}−{x}^{\mathrm{2}} =\mathrm{2}\Rightarrow{x}=\pm\mathrm{1} \\ $$
Commented by 1549442205PVT last updated on 11/Nov/21
Thank you Sir Sindhi for your shortest way of calculation
$${Thank}\:{you}\:{Sir}\:{Sindhi}\:{for}\:{your}\:{shortest}\:\: \\ $$$${way}\:{of}\:{calculation} \\ $$
Commented by Rasheed.Sindhi last updated on 11/Nov/21
You′re Welcome sir! You′re also an old forum-friend.I remember your old valueable posts.
$$\mathbb{Y}\mathrm{ou}'\mathrm{re}\:\mathbb{W}\mathrm{elcome}\:\mathrm{sir}! \\ $$$$\mathbb{Y}\mathrm{ou}'\mathrm{re}\:\mathrm{also}\:\mathrm{an}\:\mathrm{old}\:\mathrm{forum}-\mathrm{friend}.\mathbb{I} \\ $$$$\mathrm{remember}\:\mathrm{your}\:\mathrm{old}\:\mathrm{valueable}\:\mathrm{posts}. \\ $$
Commented by akolade last updated on 12/Nov/21
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 10/Nov/21
((√3) +x_(a) )^(12) +((√3) −x_(b) )^(12) =172928; x=? Simplifying LHS: a+b=(√3) +x+(√3) −x=2(√3) ab=((√3) )^2 −(x)^2 =3−x^2 a^3 +b^3 =(a+b)^3 −3ab(a+b) =(2(√3) )^3 −3(3−x^2 )(2(√3) ) =2(√3) (12−9+3x^2 )=2(√3) (3+3x^2 ) =6(√3) (1+x^2 ) a^6 +b^6 =(a^3 +b^3 )^2 −2(ab)^3 =(6(√3) (1+x^2 ))^2 −2(3−x^2 )^3 =108(1+x^2 )^2 −2(3−x^2 )^3 =2x^6 +90x^4 +270x^2 +54 ⋮ a^(12) +b^(12) =(a^6 +b^6 )^2 −2(ab)^6 ⋮ Solving the equation: ▶2x^(12) +396x^(10) +8910x^8 +49896x^6 +80190x^4 +32076x^2 +1458 =172928 ▶2x^(12) +396x^(10) +8910x^8 +49896x^6 +80190x^4 +32076x^2 −171470=0 ▶x^(12) +198x^(10) +4455x^8 +24948x^6 +40095x^4 +16038x^2 −85735=0 With the help of calculator x=±1
$$\left(\underset{{a}} {\underbrace{\sqrt{\mathrm{3}}\:+{x}}}\right)^{\mathrm{12}} +\left(\underset{{b}} {\underbrace{\sqrt{\mathrm{3}}\:−{x}}}\right)^{\mathrm{12}} =\mathrm{172928};\:{x}=? \\ $$$$\mathrm{Simplifying}\:\mathrm{LHS}: \\ $$$${a}+{b}=\sqrt{\mathrm{3}}\:+{x}+\sqrt{\mathrm{3}}\:−{x}=\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$${ab}=\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} −\left({x}\right)^{\mathrm{2}} =\mathrm{3}−{x}^{\mathrm{2}} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{3}{ab}\left({a}+{b}\right) \\ $$$$\:\:\:\:\:=\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)\left(\mathrm{2}\sqrt{\mathrm{3}}\:\right) \\ $$$$\:\:\:\:=\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{12}−\mathrm{9}+\mathrm{3}{x}^{\mathrm{2}} \right)=\mathrm{2}\sqrt{\mathrm{3}}\:\left(\mathrm{3}+\mathrm{3}{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\mathrm{6}\sqrt{\mathrm{3}}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:=\left(\mathrm{6}\sqrt{\mathrm{3}}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:\:\:=\mathrm{108}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\:=\mathrm{2}{x}^{\mathrm{6}} +\mathrm{90}{x}^{\mathrm{4}} +\mathrm{270}{x}^{\mathrm{2}} +\mathrm{54} \\ $$$$\:\:\:\:\vdots \\ $$$${a}^{\mathrm{12}} +{b}^{\mathrm{12}} =\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} \right)^{\mathrm{2}} −\mathrm{2}\left({ab}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:\vdots \\ $$$$\mathrm{Solving}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\blacktriangleright\mathrm{2}{x}^{\mathrm{12}} +\mathrm{396}{x}^{\mathrm{10}} +\mathrm{8910}{x}^{\mathrm{8}} +\mathrm{49896}{x}^{\mathrm{6}} \\ $$$$\:\:\:\:+\mathrm{80190}{x}^{\mathrm{4}} +\mathrm{32076}{x}^{\mathrm{2}} +\mathrm{1458} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{172928} \\ $$$$\blacktriangleright\mathrm{2}{x}^{\mathrm{12}} +\mathrm{396}{x}^{\mathrm{10}} +\mathrm{8910}{x}^{\mathrm{8}} +\mathrm{49896}{x}^{\mathrm{6}} \\ $$$$\:\:\:\:+\mathrm{80190}{x}^{\mathrm{4}} +\mathrm{32076}{x}^{\mathrm{2}} −\mathrm{171470}=\mathrm{0} \\ $$$$\blacktriangleright{x}^{\mathrm{12}} +\mathrm{198}{x}^{\mathrm{10}} +\mathrm{4455}{x}^{\mathrm{8}} +\mathrm{24948}{x}^{\mathrm{6}} \\ $$$$\:\:\:\:+\mathrm{40095}{x}^{\mathrm{4}} +\mathrm{16038}{x}^{\mathrm{2}} −\mathrm{85735}=\mathrm{0} \\ $$$${With}\:{the}\:{help}\:{of}\:{calculator} \\ $$$$\:\:\:\:\:{x}=\pm\mathrm{1} \\ $$
Commented by 1549442205PVT last updated on 11/Nov/21
x^(12) +198x^(10) +4455x^8 +24948x^6 +40095x^4 +16038x^2 −85735=0 ⇔(x^2 −1)(x^(10) +199x^8 +4654x^6 +29602x^4 +69697x^2 +85735)=0 ⇔x^2 =1!As this is bettter,Sir Shindhi
$$\:\:\:\:{x}^{\mathrm{12}} +\mathrm{198}{x}^{\mathrm{10}} +\mathrm{4455}{x}^{\mathrm{8}} +\mathrm{24948}{x}^{\mathrm{6}} +\mathrm{40095}{x}^{\mathrm{4}} +\mathrm{16038}{x}^{\mathrm{2}} −\mathrm{85735}=\mathrm{0} \\ $$$$\Leftrightarrow\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{10}} +\mathrm{199}{x}^{\mathrm{8}} +\mathrm{4654}{x}^{\mathrm{6}} +\mathrm{29602}{x}^{\mathrm{4}} +\mathrm{69697}{x}^{\mathrm{2}} +\mathrm{85735}\right)=\mathrm{0} \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} =\mathrm{1}!{As}\:{this}\:{is}\:{bettter},{Sir}\:{Shindhi} \\ $$
Commented by Rasheed.Sindhi last updated on 11/Nov/21
ThanX sir PVT for complement. This is even important for me because I don′t expect any feedback from the questioner. Your aproach was also good although somewhat more labarious. (sir my name is Sindhi not Shindhi)
$$\mathcal{T}{han}\mathcal{X}\:{sir}\:{PVT}\:{for}\:{complement}. \\ $$$$\mathcal{T}{his}\:{is}\:{even}\:{important}\:{for}\:{me}\:{because} \\ $$$${I}\:{don}'{t}\:{expect}\:{any}\:{feedback}\:{from}\:{the} \\ $$$${questioner}. \\ $$$${Your}\:{aproach}\:{was}\:{also}\:{good}\:{although} \\ $$$${somewhat}\:{more}\:{labarious}. \\ $$$$\left({sir}\:{my}\:{name}\:{is}\:{Sindhi}\:{not}\:{Shindhi}\right) \\ $$

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