Question Number 179371 by mathlove last updated on 28/Oct/22
Commented by Frix last updated on 28/Oct/22
$$\frac{\mathrm{1051}}{\mathrm{225}} \\ $$
Commented by mathlove last updated on 28/Oct/22
$${solve}??? \\ $$
Answered by mr W last updated on 29/Oct/22
$$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\left({using}\:{Newton}'{s}\:{Identities}\right) \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{1} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}=\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} −{e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \:\Rightarrow{e}_{\mathrm{3}} ={e}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}=\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{3}}−\mathrm{2}{e}_{\mathrm{2}} +\left({e}_{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\mathrm{2}{e}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{e}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow{e}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$${p}_{\mathrm{6}} ={e}_{\mathrm{1}} {p}_{\mathrm{5}} −{e}_{\mathrm{2}} {p}_{\mathrm{4}} +{e}_{\mathrm{3}} {p}_{\mathrm{3}} \\ $$$$\:\:\:\:=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{2}}{\mathrm{25}}+\frac{\mathrm{7}}{\mathrm{3}}\right)+\frac{\mathrm{2}}{\mathrm{15}}×\mathrm{2}=\frac{\mathrm{1406}}{\mathrm{375}} \\ $$$${p}_{\mathrm{7}} ={e}_{\mathrm{1}} {p}_{\mathrm{6}} −{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} \\ $$$$\:\:\:\:=\frac{\mathrm{1406}}{\mathrm{375}}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{15}}\left(\frac{\mathrm{2}}{\mathrm{25}}+\frac{\mathrm{7}}{\mathrm{3}}\right)=\frac{\mathrm{1051}}{\mathrm{225}}\:\checkmark \\ $$
Commented by mathlove last updated on 29/Oct/22
$${whats}\:{e}_{\mathrm{1}} \:\:{e}_{\mathrm{2}\:\:\:\:} {e}_{\mathrm{3}} \\ $$$${and}\:{p}_{\mathrm{1}} \:\:{p}_{\mathrm{2}} \:\:{p}_{\mathrm{3}} \\ $$
Commented by mr W last updated on 29/Oct/22
$${p}_{{n}} ={x}^{{n}} +{y}^{{n}} +{z}^{{n}} \\ $$$${e}_{\mathrm{1}} ={x}+{y}+{z} \\ $$$${e}_{\mathrm{2}} ={xy}+{yz}+{zx} \\ $$$${e}_{\mathrm{3}} ={xyz} \\ $$
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
$${thanks} \\ $$
Answered by mr W last updated on 29/Oct/22
$$\boldsymbol{{Method}}\:\boldsymbol{{II}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$${let}\:{s}={xy}+{yz}+{zx},\:{t}={xyz} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{1}−\mathrm{2}{s} \\ $$$$\left({xy}+{yz}+{zx}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}\left({xyz}\right)\left({x}+{y}+{z}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{2}{t} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} +\mathrm{3}\left({x}+{y}+{z}\right)\left({xy}+{yz}+{zx}\right)−\mathrm{3}{xyz} \\ $$$$\mathrm{1}=\mathrm{2}+\mathrm{3}{s}−\mathrm{3}{t} \\ $$$$\Rightarrow{t}={s}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)={x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} +{x}^{\mathrm{2}} {y}^{\mathrm{3}} +{x}^{\mathrm{3}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{3}} +{y}^{\mathrm{3}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{3}} +{z}^{\mathrm{3}} {x}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\mathrm{2}{s}\right)\mathrm{2}=\mathrm{3}+\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}+{y}+{z}\right)−{xyz}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}−\mathrm{4}{s}=\mathrm{3}+{s}^{\mathrm{2}} −\mathrm{2}{s}−\frac{\mathrm{2}}{\mathrm{3}}−{s}\left({s}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow{s}=−\frac{\mathrm{1}}{\mathrm{5}}\: \\ $$$$\Rightarrow{t}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{15}} \\ $$$$\left({x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +{x}^{\mathrm{2}} {y}^{\mathrm{5}} +{x}^{\mathrm{5}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{5}} +{y}^{\mathrm{5}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{5}} +{z}^{\mathrm{5}} {x}^{\mathrm{2}} \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2}{s}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} {y}^{\mathrm{2}} {z}^{\mathrm{2}} \left({x}+{y}+{z}\right) \\ $$$$\mathrm{3}\left(\mathrm{1}−\mathrm{2}{s}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\mathrm{2}\left({s}^{\mathrm{2}} −\mathrm{2}{t}\right)−{t}^{\mathrm{2}} \\ $$$$\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{5}}\right)={x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{25}}−\frac{\mathrm{4}}{\mathrm{15}}\right)−\frac{\mathrm{4}}{\mathrm{15}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} =\frac{\mathrm{1051}}{\mathrm{225}}\:\checkmark \\ $$
Commented by mathlove last updated on 29/Oct/22
$${thanks}\:{mr}\:{W} \\ $$