Question Number 7246 by Tawakalitu. last updated on 18/Aug/16
Answered by Yozzia last updated on 19/Aug/16
$${Define}\:{the}\:{number}\:{k}\:{such}\:{that}\: \\ $$$${k}={x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right) \\ $$$$\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$${x},{y},{z},{n}\in\mathbb{C}\:{and}\:{x}+{y}+{z}=\mathrm{0}. \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} +{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right. \\ $$$$\left.\:\:\:\:\:\:\:+{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right\} \\ $$$$ \\ $$$${Also},\:−{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)=−{x}^{{n}+\mathrm{1}} {yz}−{xy}^{{n}+\mathrm{1}} {z}−{xyz}^{{n}+\mathrm{1}} \\ $$$$ \\ $$$$\therefore\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} \right)−{x}^{{n}+\mathrm{1}} {yz}−{xy}^{{n}+\mathrm{1}} {z}−{xyz}^{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right\} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{zy}\right)−{y}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{zx}\right)−{z}^{{n}+\mathrm{1}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}\right)\right\} \\ $$$${k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left({y}+{z}\right)^{\mathrm{2}} −{y}^{{n}+\mathrm{1}} \left({x}+{z}\right)^{\mathrm{2}} −{z}^{{n}+\mathrm{1}} \left({x}+{y}\right)^{\mathrm{2}} \right\} \\ $$$${Since}\:{x}+{y}+{z}=\mathrm{0}\Rightarrow\:{x}+{y}=−{z},\:{x}+{z}=−{y},\:{y}+{z}=−{x}. \\ $$$$\therefore\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{1}} \left(−{x}\right)^{\mathrm{2}} −{y}^{{n}+\mathrm{1}} \left(−{y}\right)^{\mathrm{2}} −{z}^{{n}+\mathrm{1}} \left(−{z}\right)^{\mathrm{2}} \right\} \\ $$$$\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{x}^{{n}+\mathrm{3}} −{y}^{{n}+\mathrm{3}} −{z}^{{n}+\mathrm{3}} \right\} \\ $$$${k}=\mathrm{0},\:{if}\:{x}+{y}+{z}=\mathrm{0}. \\ $$$${Thus}\:{x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} −{xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right)=\mathrm{0} \\ $$$${or}\: \\ $$$${x}^{{n}+\mathrm{3}} +{y}^{{n}+\mathrm{3}} +{z}^{{n}+\mathrm{3}} ={xyz}\left({x}^{{n}} +{y}^{{n}} +{z}^{{n}} \right)+\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\left({x}^{{n}+\mathrm{1}} +{y}^{{n}+\mathrm{1}} +{z}^{{n}+\mathrm{1}} \right) \\ $$$${if}\:{x}+{y}+{z}=\mathrm{0}. \\ $$
Commented by Tawakalitu. last updated on 19/Aug/16
$${Wow}\:{thanks}\:{so}\:{muvh}\:{sir}.\:{i}\:{really}\:{appreciate}. \\ $$