Question-160482

Question Number 160482 by mr W last updated on 30/Nov/21

Commented by mr W last updated on 30/Nov/21

$${solve}\:{for}\:\mathbb{R} \\ $$

Answered by mr W last updated on 30/Nov/21

Using Newton′s Identities p_k =x^n +y^n +z^n e_1 =p_1 =x+y+z=3 e_2 =xy+yz+zx e_3 =xyz p_2 =e_1 p_1 −2e_2 =9−2e_2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 15=3(9−2e_2 )−3e_2 +3e_3 ⇒e_3 =3e_2 −4 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =3×15−e_2 (9−2e_2 )+3e_3 =33+2e_2 ^2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 83=3(33+2e_2 ^2 )−15e_2 +(3e_2 −4)(9−2e_2 ) e_2 =1 e_3 =3×1−4=−1 i.e. x+y+z=3 xy+yz+zx=1 xyz=−1 so x,y,z are roots of t^3 −3t^2 +t+1=0 (t−1)(t^2 −2t−1)=0 t=1, 1+(√2), 1−(√2) ⇒(x,y,z)=(1,1+(√2),1−(√2))

$${Using}\:{Newton}'{s}\:{Identities} \\ $$$${p}_{{k}} ={x}^{{n}} +{y}^{{n}} +{z}^{{n}} \\ $$$${e}_{\mathrm{1}} ={p}_{\mathrm{1}} ={x}+{y}+{z}=\mathrm{3} \\ $$$${e}_{\mathrm{2}} ={xy}+{yz}+{zx} \\ $$$${e}_{\mathrm{3}} ={xyz} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{9}−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\mathrm{15}=\mathrm{3}\left(\mathrm{9}−\mathrm{2}{e}_{\mathrm{2}} \right)−\mathrm{3}{e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow{e}_{\mathrm{3}} =\mathrm{3}{e}_{\mathrm{2}} −\mathrm{4} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\mathrm{3}×\mathrm{15}−{e}_{\mathrm{2}} \left(\mathrm{9}−\mathrm{2}{e}_{\mathrm{2}} \right)+\mathrm{3}{e}_{\mathrm{3}} =\mathrm{33}+\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} \\ $$$$\mathrm{83}=\mathrm{3}\left(\mathrm{33}+\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} \right)−\mathrm{15}{e}_{\mathrm{2}} +\left(\mathrm{3}{e}_{\mathrm{2}} −\mathrm{4}\right)\left(\mathrm{9}−\mathrm{2}{e}_{\mathrm{2}} \right) \\ $$$${e}_{\mathrm{2}} =\mathrm{1} \\ $$$${e}_{\mathrm{3}} =\mathrm{3}×\mathrm{1}−\mathrm{4}=−\mathrm{1} \\ $$$${i}.{e}. \\ $$$${x}+{y}+{z}=\mathrm{3} \\ $$$${xy}+{yz}+{zx}=\mathrm{1} \\ $$$${xyz}=−\mathrm{1} \\ $$$${so}\:{x},{y},{z}\:{are}\:{roots}\:{of}\:{t}^{\mathrm{3}} −\mathrm{3}{t}^{\mathrm{2}} +{t}+\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{1},\:\mathrm{1}+\sqrt{\mathrm{2}},\:\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\mathrm{1},\mathrm{1}+\sqrt{\mathrm{2}},\mathrm{1}−\sqrt{\mathrm{2}}\right) \\ $$

Commented by Tawa11 last updated on 30/Nov/21

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by MJS_new last updated on 30/Nov/21

(1) z=3−x−y let x=u−v∧y=u+v ⇒ (2) v^2 =((u^3 −6u^2 +9u−2)/u) (3) v^4 +2u^2 v^2 −((3u^5 −24u^4 +72u^3 −108u^2 +81u−16)/u)=0 (2) in (3) ⇒ u^3 −3u^2 +(5/2)u−(1/2)=0 u_1 =1 u_(2, 3) =1±((√2)/2) the rest is easy

$$\left(\mathrm{1}\right)\:{z}=\mathrm{3}−{x}−{y} \\ $$$$\mathrm{let}\:{x}={u}−{v}\wedge{y}={u}+{v} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}\right)\:{v}^{\mathrm{2}} =\frac{{u}^{\mathrm{3}} −\mathrm{6}{u}^{\mathrm{2}} +\mathrm{9}{u}−\mathrm{2}}{{u}} \\ $$$$\left(\mathrm{3}\right)\:{v}^{\mathrm{4}} +\mathrm{2}{u}^{\mathrm{2}} {v}^{\mathrm{2}} −\frac{\mathrm{3}{u}^{\mathrm{5}} −\mathrm{24}{u}^{\mathrm{4}} +\mathrm{72}{u}^{\mathrm{3}} −\mathrm{108}{u}^{\mathrm{2}} +\mathrm{81}{u}−\mathrm{16}}{{u}}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{in}\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow \\ $$$${u}^{\mathrm{3}} −\mathrm{3}{u}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}{u}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$${u}_{\mathrm{1}} =\mathrm{1} \\ $$$${u}_{\mathrm{2},\:\mathrm{3}} =\mathrm{1}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$

Commented by mr W last updated on 30/Nov/21

$${nice}\:{approach}! \\ $$

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