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Question Number 179371 by mathlove last updated on 28/Oct/22
Commented by Frix last updated on 28/Oct/22
((1051)/(225))
$$\frac{\mathrm{1051}}{\mathrm{225}} \\ $$
Commented by mathlove last updated on 28/Oct/22
solve???
$${solve}??? \\ $$
Answered by mr W last updated on 29/Oct/22
Method I (using Newton′s Identities) p_1 =e_1 =1 p_2 =e_1 p_1 −2e_2 =1−2e_2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 ⇒2=1−2e_2 −e_2 +3e_3 ⇒e_3 =e_2 +(1/3) p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =2e_2 ^2 +(7/3) p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 ⇒3=2e_2 ^2 +(7/3)−2e_2 +(e_2 +(1/3))(1−2e_2 ) ⇒e_2 =−(1/5) ⇒e_3 =−(1/5)+(1/3)=(2/(15)) p_6 =e_1 p_5 −e_2 p_4 +e_3 p_3 =3+(1/5)((2/(25))+(7/3))+(2/(15))×2=((1406)/(375)) p_7 =e_1 p_6 −e_2 p_5 +e_3 p_4 =((1406)/(375))+(3/5)+(2/(15))((2/(25))+(7/3))=((1051)/(225)) ✓
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\left({using}\:{Newton}'{s}\:{Identities}\right) \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}=\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} −{e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow{e}_{\mathrm{3}} ={e}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}=\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}}−\mathrm{2}{e}_{\mathrm{2}} +\left({e}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow{e}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$${p}_{\mathrm{6}} ={e}_{\mathrm{1}} {p}_{\mathrm{5}} −{e}_{\mathrm{2}} {p}_{\mathrm{4}} +{e}_{\mathrm{3}} {p}_{\mathrm{3}} \\ $$$$\:\:\:\:=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{2}}{\mathrm{25}}+\frac{\mathrm{7}}{\mathrm{3}}\right)+\frac{\mathrm{2}}{\mathrm{15}}×\mathrm{2}=\frac{\mathrm{1406}}{\mathrm{375}} \\ $$$${p}_{\mathrm{7}} ={e}_{\mathrm{1}} {p}_{\mathrm{6}} −{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{1406}}{\mathrm{375}}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{15}}\left(\frac{\mathrm{2}}{\mathrm{25}}+\frac{\mathrm{7}}{\mathrm{3}}\right)=\frac{\mathrm{1051}}{\mathrm{225}}\:\checkmark \\ $$
Commented by mathlove last updated on 29/Oct/22
whats e_1 e_(2 ) e_3 and p_1 p_2 p_3
$${whats}\:{e}_{\mathrm{1}} \:\:{e}_{\mathrm{2}\:\:\:\:} {e}_{\mathrm{3}} \\ $$$${and}\:{p}_{\mathrm{1}} \:\:{p}_{\mathrm{2}} \:\:{p}_{\mathrm{3}} \\ $$
Commented by mr W last updated on 29/Oct/22
p_n =x^n +y^n +z^n e_1 =x+y+z e_2 =xy+yz+zx e_3 =xyz
$${p}_{{n}} ={x}^{{n}} +{y}^{{n}} +{z}^{{n}} \\ $$$${e}_{\mathrm{1}} ={x}+{y}+{z} \\ $$$${e}_{\mathrm{2}} ={xy}+{yz}+{zx} \\ $$$${e}_{\mathrm{3}} ={xyz} \\ $$
Commented by mr W last updated on 29/Oct/22
Commented by mr W last updated on 29/Oct/22
https://en.m.wikipedia.org/wiki/Newton%27s_identities
Commented by mathlove last updated on 29/Oct/22
thanks
$${thanks} \\ $$
Answered by mr W last updated on 29/Oct/22
Method II (x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz+zx) let s=xy+yz+zx, t=xyz ⇒x^2 +y^2 +z^2 =1−2s (xy+yz+zx)^2 =x^2 y^2 +y^2 z^2 +z^2 x^2 +2(xyz)(x+y+z) ⇒x^2 y^2 +y^2 z^2 +z^2 x^2 =s^2 −2t (x+y+z)^3 =x^3 +y^3 +z^3 +3(x+y+z)(xy+yz+zx)−3xyz 1=2+3s−3t ⇒t=s+(1/3) (x^2 +y^2 +z^2 )(x^3 +y^3 +z^3 )=x^5 +y^5 +z^5 +x^2 y^3 +x^3 y^2 +y^2 z^3 +y^3 z^2 +z^2 x^3 +z^3 x^2 (1−2s)2=3+(x^2 y^2 +y^2 z^2 +z^2 x^2 )(x+y+z)−xyz(xy+yz+zx) 2−4s=3+s^2 −2s−(2/3)−s(s+(1/3)) ⇒s=−(1/5) ⇒t=−(1/5)+(1/3)=(2/(15)) (x^5 +y^5 +z^5 )(x^2 +y^2 +z^2 )=x^7 +y^7 +z^7 +x^2 y^5 +x^5 y^2 +y^2 z^5 +y^5 z^2 +z^2 x^5 +z^5 x^2 3(1−2s)=x^7 +y^7 +z^7 +(x^3 +y^3 +z^3 )(x^2 y^2 +y^2 z^2 +z^2 x^2 )−x^2 y^2 z^2 (x+y+z) 3(1−2s)=x^7 +y^7 +z^7 +2(s^2 −2t)−t^2 3(1+(2/5))=x^7 +y^7 +z^7 +2((1/(25))−(4/(15)))−(4/(15^2 )) ⇒x^7 +y^7 +z^7 =((1051)/(225)) ✓
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$${let}\:{s}={xy}+{yz}+{zx},\:{t}={xyz} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{s} \\ $$$$\left({xy}+{yz}+{zx}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}\left({xyz}\right)\left({x}+{y}+{z}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{2}{t} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)−\mathrm{3}{xyz} \\ $$$$\mathrm{1}=\mathrm{2}+\mathrm{3}{s}−\mathrm{3}{t} \\ $$$$\Rightarrow{t}={s}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)={x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}^{\mathrm{3}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{3}} +{y}^{\mathrm{3}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{3}} +{z}^{\mathrm{3}} {x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{s}\right)\mathrm{2}=\mathrm{3}+\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}+{y}+{z}\right)−{xyz}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}−\mathrm{4}{s}=\mathrm{3}+{s}^{\mathrm{2}} −\mathrm{2}{s}−\frac{\mathrm{2}}{\mathrm{3}}−{s}\left({s}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow{s}=−\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow{t}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$$\left({x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +{x}^{\mathrm{2}} {y}^{\mathrm{5}} +{x}^{\mathrm{5}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{5}} +{y}^{\mathrm{5}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{5}} +{z}^{\mathrm{5}} {x}^{\mathrm{2}} \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2}{s}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \left({x}+{y}+{z}\right) \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2}{s}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\mathrm{2}\left({s}^{\mathrm{2}} −\mathrm{2}{t}\right)−{t}^{\mathrm{2}} \\ $$$$\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{5}}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{25}}−\frac{\mathrm{4}}{\mathrm{15}}\right)−\frac{\mathrm{4}}{\mathrm{15}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} =\frac{\mathrm{1051}}{\mathrm{225}}\:\checkmark \\ $$
Commented by mathlove last updated on 29/Oct/22
thanks mr W
$${thanks}\:{mr}\:{W} \\ $$

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