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Question Number 178476 by Shrinava last updated on 17/Oct/22

Find a∈R Such that x_1 ^(16) + x_2 ^(16) + x_3 ^(16) = 30 Where x_1 ,x_2 ,x_3 − are the roots of the equation: x^3 + ax + 1 = 0

$$\mathrm{Find}\:\:\mathrm{a}\in\mathbb{R} \\ $$$$\mathrm{Such}\:\mathrm{that}\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{16}} \:+\:\mathrm{x}_{\mathrm{2}} ^{\mathrm{16}} \:+\:\mathrm{x}_{\mathrm{3}} ^{\mathrm{16}} \:=\:\mathrm{30} \\ $$$$\mathrm{Where}\:\:\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,\mathrm{x}_{\mathrm{3}} −\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{ax}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$

Answered by Frix last updated on 17/Oct/22

x^3 +ax+1=0 ⇒ x_k =((2(√(−3a)))/3)sin ((2πk+sin^(−1) ((3(√3))/(2(−a)^(3/2) )))/3) with k=1, 2, 3 let α=((3(√3))/(2(−a)^(3/2) )) x_k =((2(√(−3a)))/3)sin ((2πk+sin^(−1) α)/3) x_k ^(16) =((65536a^8 )/(6561))sin^(16) ((2πk+sin^(−1) α)/3) sin ((2π+sin^(−1) α)/3) =(((√3)cos ((sin^(−1) α)/3) −sin ((sin^(−1) α)/3))/2)=(((√3)c−s)/2) sin ((4π+sin^(−1) α)/3) =−(((√3)cos ((sin^(−1) α)/3) +sin ((sin^(−1) α)/3))/2)=−(((√3)c+s)/2) sin ((6π+sin^(−1) α)/3) =sin ((sin^(−1) α)/3) =s ((((√3)c−s)/2))^(16) +(−(((√3)c+s)/2))^(16) +s^(16) = (with c=(√(1−s^2 ))) =((45s^(12) )/2)−((135s^(10) )/2)+((1215s^8 )/(16))−((1701s^6 )/(64))−((5103s^4 )/(512))+((6561s^2 )/(1024))+((6561)/(32768))= (after some work) =((360cos (12×((sin^(−1) α)/3)) −13104cos (6×((sin^(−1) α)/3)) +19305)/(32768))= =((45α^4 )/(512))+((729α^2 )/(1024))+((6561)/(32768))= =((6561)/(32768))−((19683)/(4086a^3 ))+((32805)/(8192a^6 )) multiplied with ((65536a^8 )/(6561)) x_1 ^(16) +x_2 ^(16) +x_3 ^(16) =2a^8 −48a^5 +40a^2 =30 we need to solve a^8 −24a^5 +20a^2 −15=0 I can only approximate a_1 ≈−.718638134 a_2 ≈2.85282288 no other real solutions

$${x}^{\mathrm{3}} +{ax}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}_{{k}} =\frac{\mathrm{2}\sqrt{−\mathrm{3}{a}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi{k}+\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\left(−{a}\right)^{\mathrm{3}/\mathrm{2}} }}{\mathrm{3}}\:\mathrm{with}\:{k}=\mathrm{1},\:\mathrm{2},\:\mathrm{3} \\ $$$$\mathrm{let}\:\alpha=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}\left(−{a}\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${x}_{{k}} =\frac{\mathrm{2}\sqrt{−\mathrm{3}{a}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{2}\pi{k}+\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}} \\ $$$${x}_{{k}} ^{\mathrm{16}} =\frac{\mathrm{65536}{a}^{\mathrm{8}} }{\mathrm{6561}}\mathrm{sin}^{\mathrm{16}} \:\frac{\mathrm{2}\pi{k}+\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}} \\ $$$$\mathrm{sin}\:\frac{\mathrm{2}\pi+\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:=\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:−\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{c}−{s}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{\mathrm{4}\pi+\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:=−\frac{\sqrt{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:+\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}}{\mathrm{2}}=−\frac{\sqrt{\mathrm{3}}{c}+{s}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{\mathrm{6}\pi+\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:=\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\:={s} \\ $$$$\left(\frac{\sqrt{\mathrm{3}}{c}−{s}}{\mathrm{2}}\right)^{\mathrm{16}} +\left(−\frac{\sqrt{\mathrm{3}}{c}+{s}}{\mathrm{2}}\right)^{\mathrm{16}} +{s}^{\mathrm{16}} = \\ $$$$\left(\mathrm{with}\:{c}=\sqrt{\mathrm{1}−{s}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{45}{s}^{\mathrm{12}} }{\mathrm{2}}−\frac{\mathrm{135}{s}^{\mathrm{10}} }{\mathrm{2}}+\frac{\mathrm{1215}{s}^{\mathrm{8}} }{\mathrm{16}}−\frac{\mathrm{1701}{s}^{\mathrm{6}} }{\mathrm{64}}−\frac{\mathrm{5103}{s}^{\mathrm{4}} }{\mathrm{512}}+\frac{\mathrm{6561}{s}^{\mathrm{2}} }{\mathrm{1024}}+\frac{\mathrm{6561}}{\mathrm{32768}}= \\ $$$$\left(\mathrm{after}\:\mathrm{some}\:\mathrm{work}\right) \\ $$$$=\frac{\mathrm{360cos}\:\left(\mathrm{12}×\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\right)\:−\mathrm{13104cos}\:\left(\mathrm{6}×\frac{\mathrm{sin}^{−\mathrm{1}} \:\alpha}{\mathrm{3}}\right)\:+\mathrm{19305}}{\mathrm{32768}}= \\ $$$$=\frac{\mathrm{45}\alpha^{\mathrm{4}} }{\mathrm{512}}+\frac{\mathrm{729}\alpha^{\mathrm{2}} }{\mathrm{1024}}+\frac{\mathrm{6561}}{\mathrm{32768}}= \\ $$$$=\frac{\mathrm{6561}}{\mathrm{32768}}−\frac{\mathrm{19683}}{\mathrm{4086}{a}^{\mathrm{3}} }+\frac{\mathrm{32805}}{\mathrm{8192}{a}^{\mathrm{6}} } \\ $$$$\mathrm{multiplied}\:\mathrm{with}\:\frac{\mathrm{65536}{a}^{\mathrm{8}} }{\mathrm{6561}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{16}} +{x}_{\mathrm{2}} ^{\mathrm{16}} +{x}_{\mathrm{3}} ^{\mathrm{16}} =\mathrm{2}{a}^{\mathrm{8}} −\mathrm{48}{a}^{\mathrm{5}} +\mathrm{40}{a}^{\mathrm{2}} =\mathrm{30} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{solve} \\ $$$${a}^{\mathrm{8}} −\mathrm{24}{a}^{\mathrm{5}} +\mathrm{20}{a}^{\mathrm{2}} −\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${a}_{\mathrm{1}} \approx−.\mathrm{718638134} \\ $$$${a}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{85282288} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solutions} \\ $$

Commented by Shrinava last updated on 17/Oct/22

Sorry professor, x_1 ^(16) +x_2 ^(16) +x_3 ^(16) =90

$$\mathrm{Sorry}\:\mathrm{professor},\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{16}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{16}} +\mathrm{x}_{\mathrm{3}} ^{\mathrm{16}} =\mathrm{90} \\ $$

Commented by Ar Brandon last updated on 17/Oct/22

Greetings, old man �� I already began to feel your absence and was missing you already, dear Sir�� Little did I know that...��

Commented by Frix last updated on 17/Oct/22

x_1 ^(16) +x_2 ^(16) +x_3 ^(16) =2a^8 −48a^5 +40a^2 =90 a^8 −24a^5 +20a^2 −45=0 a_1 =−1 a_2 ≈2.85944202 no other real solutions

$${x}_{\mathrm{1}} ^{\mathrm{16}} +{x}_{\mathrm{2}} ^{\mathrm{16}} +{x}_{\mathrm{3}} ^{\mathrm{16}} =\mathrm{2}{a}^{\mathrm{8}} −\mathrm{48}{a}^{\mathrm{5}} +\mathrm{40}{a}^{\mathrm{2}} =\mathrm{90} \\ $$$${a}^{\mathrm{8}} −\mathrm{24}{a}^{\mathrm{5}} +\mathrm{20}{a}^{\mathrm{2}} −\mathrm{45}=\mathrm{0} \\ $$$${a}_{\mathrm{1}} =−\mathrm{1} \\ $$$${a}_{\mathrm{2}} \approx\mathrm{2}.\mathrm{85944202} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{real}\:\mathrm{solutions} \\ $$

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