Question Number 222404 by ajfour last updated on 25/Jun/25
Answered by ajfour last updated on 26/Jun/25
$${x}+{y}+{z}=\boldsymbol{{s}} \\ $$$${xy}+{yz}+{zx}=\boldsymbol{{k}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\boldsymbol{{h}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\boldsymbol{{A}} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\boldsymbol{{u}} \\ $$$${xyz}={t} \\ $$$${s}^{\mathrm{2}} ={h}+\mathrm{2}{k}\:\:\:\:..\left(\mathrm{1}\right) \\ $$$$\mathrm{2}{h}+\lambda{k}={A}\:\:\:\:\:…\left(\mathrm{2}\right) \\ $$$$\mathrm{2}{s}^{\mathrm{2}} ={A}−\lambda{k}+\mathrm{4}{k} \\ $$$${k}=\frac{{A}−\mathrm{2}{s}^{\mathrm{2}} }{\lambda−\mathrm{4}}\:\:\:\:\:\:\:\:\left({i}\right) \\ $$$$\lambda{s}^{\mathrm{2}} =\lambda{h}+\mathrm{2}{A}−\mathrm{4}{h} \\ $$$${h}=\frac{\lambda{s}^{\mathrm{2}} −\mathrm{2}{A}}{\lambda−\mathrm{4}}\:\:\:\:\:\:\left({ii}\right) \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{3}} ={s}^{\mathrm{3}} ={u}+\Sigma\mathrm{3}{xy}\left({x}+{y}\right)+\mathrm{6}{xyz} \\ $$$${as}\:{x}+{y}={s}−{z} \\ $$$${s}^{\mathrm{3}} ={u}+\mathrm{3}{s}\Sigma{xy}−\mathrm{3}{t} \\ $$$${s}^{\mathrm{3}} ={u}+\mathrm{3}{sk}−\mathrm{3}{t}\:\:\:\:\:\:\:…\left(\mathrm{3}\right) \\ $$$${now} \\ $$$${sh}={u}+{sk}−\mathrm{3}{t}\:\:\:\:\:\:\:…\left(\mathrm{4}\right) \\ $$$$\blacksquare \\ $$