Question Number 203157 by navin12345 last updated on 11/Jan/24
Answered by witcher3 last updated on 11/Jan/24
$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{2x}−\mathrm{2z}+\mathrm{2}=\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\in\left\{\left(\mathrm{1},\mathrm{0},\mathrm{1}\right)\right\} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\mathrm{z}^{\mathrm{3}} =\mathrm{2} \\ $$
Answered by BaliramKumar last updated on 11/Jan/24
$$\mathrm{1}{x}^{\mathrm{2}} +\mathrm{1}{y}^{\mathrm{2}} +\mathrm{1}{z}^{\mathrm{2}} \:=\:\mathrm{2}\left(\mathrm{1}{x}+\mathrm{0}{y}+\mathrm{1}{z}−\mathrm{1}\right) \\ $$$${x}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\:\mathrm{1} \\ $$$${y}\:=\:\frac{\mathrm{0}}{\mathrm{1}}\:=\:\mathrm{0} \\ $$$${z}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\:\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\:{y}^{\mathrm{3}} +\:{z}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$