Question Number 52592 by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19
$${if}\:{x}+{y}+{z}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} =\mathrm{3} \\ $$$${find}\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \\ $$
Answered by math1967 last updated on 10/Jan/19
$${x}+{y}+{z}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$$\therefore\left({xy}+{yz}+{zx}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{xyz}\left({x}+{y}+{z}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{12}}\:\bigstar \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} −\mathrm{3}{xyz}+\mathrm{3}{xyz}=\mathrm{3}\bigstar \\ $$$$\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −{xy}−{yz}−{zx}\right)+\mathrm{3}{xyz}=\mathrm{3} \\ $$$$\mathrm{3}{xyz}=\mathrm{3}−\frac{\mathrm{5}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{xyz}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{4} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} +\mathrm{2}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{y}^{\mathrm{2}} {z}^{\mathrm{2}} +{z}^{\mathrm{2}} {x}^{\mathrm{2}} \right)=\mathrm{4} \\ $$$${x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} =\mathrm{4}+\frac{\mathrm{1}}{\mathrm{6}}=\mathrm{4}\frac{\mathrm{1}}{\mathrm{6}}{ans} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19
$${bah}\:{darun}\:{excellent}… \\ $$
Commented by math1967 last updated on 10/Jan/19
$${dhanyabad} \\ $$