Question Number 35291 by 123 45 polytechnicien last updated on 17/May/18
$${solve}\:\:{in}\:{Z}\:\:{x}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{3}} =\mathrm{4}{z}^{\mathrm{3}} \\ $$
Commented by 123 45 polytechnicien last updated on 17/May/18
$${why}\:{z}={x}+{iy}\:{and}\:{nota}+{ib}\:? \\ $$
Answered by ajfour last updated on 17/May/18
$$\mathrm{4}\left({x}+{iy}\right)^{\mathrm{3}} ={x}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{3}} \\ $$$$\mathrm{4}\left({x}^{\mathrm{3}} −{iy}^{\mathrm{3}} +\mathrm{3}{ix}^{\mathrm{2}} {y}−\mathrm{3}{xy}^{\mathrm{2}} \right)={x}^{\mathrm{3}} +\mathrm{6}{y}^{\mathrm{3}} \\ $$$$\Rightarrow\:{y}^{\mathrm{3}} =\mathrm{3}{x}^{\mathrm{2}} {y}\:\:\:\:….\left({i}\right)\:\bar {\:}\:{and} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{xy}^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{3}} =\mathrm{0}\:\:\:\:\:….\left({ii}\right) \\ $$$$\left({i}\right)\Rightarrow\:\:{y}=\mathrm{0}\:\:{or}\:{y}^{\mathrm{2}} =\mathrm{3}{x}^{\mathrm{2}} \\ $$$${using}\:{above} \\ $$$$\left({ii}\right)\Rightarrow\:\:{x}=\mathrm{0}\:\:{or} \\ $$$$\:\:\:\:\:\:\:−\mathrm{11}{x}^{\mathrm{3}} =\mathrm{2}{y}^{\mathrm{3}} \\ $$$${or}\:\:\:\:\:\:−\mathrm{11}{x}=\mathrm{6}{y} \\ $$$${inconsistent}\:,\:{so} \\ $$$${only}\:{solution}\:{is}\:\:\left(\mathrm{0},\mathrm{0}\right)\:. \\ $$