Question Number 166009 by Rasheed.Sindhi last updated on 11/Feb/22
$$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{70}}\\{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{64}}\\{{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{2002}}\\{\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right)=?}\end{cases}\: \\ $$$$\left({Use}\:\boldsymbol{{Newton}}-\boldsymbol{{Identities}}\right. \\ $$$$\left.{or}\:{otherwise}\right) \\ $$
Commented by mr W last updated on 11/Feb/22
$${in}\:{reversed}\:{direction}\:{there}\:{are}\:{multiple} \\ $$$${solutions}. \\ $$$${the}\:{equation}\:{system}\:{has}\:\mathrm{24}\:{solution} \\ $$$${triples}\:\left({x},{y},{z}\right). \\ $$
Commented by Rasheed.Sindhi last updated on 12/Feb/22
$$\mathrm{24}!!!!!!!!!!!!!!!!!...\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{expect}! \\ $$
Commented by Rasheed.Sindhi last updated on 12/Feb/22
$$\mathrm{This}\:\mathrm{is}\:\mathrm{an}\:\mathrm{altered}\:\mathrm{form}\:\mathrm{of}\:{Q}#\mathrm{165984}. \\ $$
Commented by mr W last updated on 12/Feb/22
$$\mathrm{4}^{{th}} \:{degree}\:\Rightarrow\mathrm{4}\:{roots}. \\ $$$${symmetry}\:{for}\:{x},{y},{z}\:\Rightarrow{each}\:{roots}\:{means}\:\mathrm{6}\:{triples} \\ $$$${therefore}\:\mathrm{4}×\mathrm{6}=\mathrm{24}\:{triples} \\ $$
Answered by mr W last updated on 12/Feb/22
![S=(x+y)(y+z)(z+x) =2xyz+xy(x+y)+yz(y+z)+zx(z+x) =(x+y+z)(xy+yz+zx)−xyz =e_1 e_2 −e_3 p_2 =e_1 p_1 −2e_2 ⇒e_1 ^2 −2e_2 =70 p_3 =e_1 p_2 −e_2 p_1 +3e_3 ⇒70e_1 −e_1 e_2 +3e_3 =64 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 ⇒64e_1 −70e_2 +e_1 e_3 =2002 6e_3 =e_1 ^3 −210e_1 +128 e_1 ^4 −420e_1 ^2 +512e_1 +2688=0 (e_1 +2)(e_1 ^3 −2e_1 ^2 −416e_1 +1344)=0 ⇒e_1 =−2 ⇒e_1 ^3 −2e_1 ^2 −416e_1 +1344=0 let e_1 =s+(2/3) s^3 −((1252)/3)s+((28784)/(27))=0 s=((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313))))) ⇒e_1 =(2/3)+((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313))))) e_2 =(e_1 ^2 /2)−35 e_3 =(e_1 ^3 /6)−35e_1 +((64)/3) S=e_1 e_2 −e_3 =((e_1 ^3 −64)/3) ⇒S_1 =(((−2)^3 −64 )/3)=−24 ⇒S_(2,3,4) =(1/3){[(2/3)+((4(√(313)))/3) sin (((2kπ)/3)+(1/3) sin^(−1) ((1799)/( 313(√(313)))))]^3 −64} ≈−9.7517, 2516.6841, −3080.2657](Q166019.png)
$${S}=\left({x}+{y}\right)\left({y}+{z}\right)\left({z}+{x}\right) \\ $$$$=\mathrm{2}{xyz}+{xy}\left({x}+{y}\right)+{yz}\left({y}+{z}\right)+{zx}\left({z}+{x}\right) \\ $$$$=\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)−{xyz} \\ $$$$={e}_{\mathrm{1}} {e}_{\mathrm{2}} −{e}_{\mathrm{3}} \\ $$$$ \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} \\ $$$$\Rightarrow{e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{70} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{70}{e}_{\mathrm{1}} −{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{64} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{64}{e}_{\mathrm{1}} −\mathrm{70}{e}_{\mathrm{2}} +{e}_{\mathrm{1}} {e}_{\mathrm{3}} =\mathrm{2002} \\ $$$$ \\ $$$$\mathrm{6}{e}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{210}{e}_{\mathrm{1}} +\mathrm{128} \\ $$$${e}_{\mathrm{1}} ^{\mathrm{4}} −\mathrm{420}{e}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{512}{e}_{\mathrm{1}} +\mathrm{2688}=\mathrm{0} \\ $$$$\left({e}_{\mathrm{1}} +\mathrm{2}\right)\left({e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{2}{e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{416}{e}_{\mathrm{1}} +\mathrm{1344}\right)=\mathrm{0} \\ $$$$\Rightarrow{e}_{\mathrm{1}} =−\mathrm{2} \\ $$$$\Rightarrow{e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{2}{e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{416}{e}_{\mathrm{1}} +\mathrm{1344}=\mathrm{0} \\ $$$${let}\:{e}_{\mathrm{1}} ={s}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${s}^{\mathrm{3}} −\frac{\mathrm{1252}}{\mathrm{3}}{s}+\frac{\mathrm{28784}}{\mathrm{27}}=\mathrm{0} \\ $$$${s}=\frac{\mathrm{4}\sqrt{\mathrm{313}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1799}}{\:\mathrm{313}\sqrt{\mathrm{313}}}\right) \\ $$$$\Rightarrow{e}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{4}\sqrt{\mathrm{313}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1799}}{\:\mathrm{313}\sqrt{\mathrm{313}}}\right) \\ $$$$ \\ $$$${e}_{\mathrm{2}} =\frac{{e}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}}−\mathrm{35} \\ $$$${e}_{\mathrm{3}} =\frac{{e}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{6}}−\mathrm{35}{e}_{\mathrm{1}} +\frac{\mathrm{64}}{\mathrm{3}} \\ $$$${S}={e}_{\mathrm{1}} {e}_{\mathrm{2}} −{e}_{\mathrm{3}} =\frac{{e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{64}}{\mathrm{3}} \\ $$$$\Rightarrow{S}_{\mathrm{1}} =\frac{\left(−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{64}\:}{\mathrm{3}}=−\mathrm{24} \\ $$$$\Rightarrow{S}_{\mathrm{2},\mathrm{3},\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left\{\left[\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{4}\sqrt{\mathrm{313}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1799}}{\:\mathrm{313}\sqrt{\mathrm{313}}}\right)\right]^{\mathrm{3}} −\mathrm{64}\right\} \\ $$$$\approx−\mathrm{9}.\mathrm{7517},\:\mathrm{2516}.\mathrm{6841},\:−\mathrm{3080}.\mathrm{2657} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Feb/22
$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by nurtani last updated on 12/Feb/22
$${Good}\:{job}\:{sir}\:!!! \\ $$
Commented by Tawa11 last updated on 12/Feb/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$