Question-97675

Question Number 97675 by Farruxjano last updated on 09/Jun/20

Answered by mr W last updated on 09/Jun/20

p_n =x^n +y^n +z^n e_1 =x+y+z e_2 =xy+yz+zx e_3 =xyz e_(≥4) =0 p_1 =e_1 =0 p_2 =e_1 p_1 −2e_2 =−2e_2 p_3 =e_1 p_2 −e_2 p_1 +3e_3 =3e_3 p_4 =e_1 p_3 −e_2 p_2 +e_3 p_1 =−e_2 p_2 =2e_2 ^2 p_5 =e_1 p_4 −e_2 p_3 +e_3 p_2 =−e_2 p_3 +e_3 p_2 =−5e_2 e_3 p_7 =e_1 p_6 −e_2 p_5 +e_3 p_4 =−e_2 p_5 +e_3 p_4 =7e_2 ^2 e_3 (p_7 /7)=e_2 ^2 e_3 (p_5 /5)=−e_2 e_3 (p_2 /2)=−e_2 (p_5 /5)×(p_2 /2)=(−e_2 e_3 )(−e_2 )=e_2 ^2 e_3 =(p_7 /7) ⇒((x^7 +y^7 +z^7 )/7)=((x^5 +y^5 +z^5 )/5)×((x^2 +y^2 +z^2 )/2)

$${p}_{{n}} ={x}^{{n}} +{y}^{{n}} +{z}^{{n}} \\ $$$${e}_{\mathrm{1}} ={x}+{y}+{z} \\ $$$${e}_{\mathrm{2}} ={xy}+{yz}+{zx} \\ $$$${e}_{\mathrm{3}} ={xyz} \\ $$$${e}_{\geqslant\mathrm{4}} =\mathrm{0} \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} =\mathrm{0} \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} {p}_{\mathrm{1}} −\mathrm{2}{e}_{\mathrm{2}} =−\mathrm{2}{e}_{\mathrm{2}} \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} {p}_{\mathrm{2}} −{e}_{\mathrm{2}} {p}_{\mathrm{1}} +\mathrm{3}{e}_{\mathrm{3}} =\mathrm{3}{e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{4}} ={e}_{\mathrm{1}} {p}_{\mathrm{3}} −{e}_{\mathrm{2}} {p}_{\mathrm{2}} +{e}_{\mathrm{3}} {p}_{\mathrm{1}} =−{e}_{\mathrm{2}} {p}_{\mathrm{2}} =\mathrm{2}{e}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${p}_{\mathrm{5}} ={e}_{\mathrm{1}} {p}_{\mathrm{4}} −{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−{e}_{\mathrm{2}} {p}_{\mathrm{3}} +{e}_{\mathrm{3}} {p}_{\mathrm{2}} =−\mathrm{5}{e}_{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$${p}_{\mathrm{7}} ={e}_{\mathrm{1}} {p}_{\mathrm{6}} −{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} =−{e}_{\mathrm{2}} {p}_{\mathrm{5}} +{e}_{\mathrm{3}} {p}_{\mathrm{4}} =\mathrm{7}{e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$$\frac{{p}_{\mathrm{7}} }{\mathrm{7}}={e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$$\frac{{p}_{\mathrm{5}} }{\mathrm{5}}=−{e}_{\mathrm{2}} {e}_{\mathrm{3}} \\ $$$$\frac{{p}_{\mathrm{2}} }{\mathrm{2}}=−{e}_{\mathrm{2}} \\ $$$$\frac{{p}_{\mathrm{5}} }{\mathrm{5}}×\frac{{p}_{\mathrm{2}} }{\mathrm{2}}=\left(−{e}_{\mathrm{2}} {e}_{\mathrm{3}} \right)\left(−{e}_{\mathrm{2}} \right)={e}_{\mathrm{2}} ^{\mathrm{2}} {e}_{\mathrm{3}} =\frac{{p}_{\mathrm{7}} }{\mathrm{7}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{7}} +{y}^{\mathrm{7}} +{z}^{\mathrm{7}} }{\mathrm{7}}=\frac{{x}^{\mathrm{5}} +{y}^{\mathrm{5}} +{z}^{\mathrm{5}} }{\mathrm{5}}×\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by Farruxjano last updated on 09/Jun/20

$$\boldsymbol{{Wow}},\:\boldsymbol{{very}}\:\boldsymbol{{coooool}},\:\boldsymbol{{thanks}},\:\boldsymbol{{very}}\:\boldsymbol{{much}} \\ $$

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