Question Number 76246 by benjo last updated on 25/Dec/19
$$\mathrm{how}\:\mathrm{to}\:\mathrm{solving}\:\mathrm{x}^{\mathrm{3}} \:+\mathrm{y}^{\mathrm{3}\:} \:=\mathrm{4}\:\mathrm{and}\: \\ $$$$\mathrm{x}×\mathrm{y}\:=\mathrm{1}? \\ $$
Answered by john santuy last updated on 25/Dec/19
$${let}\:{u}\:={x}^{\mathrm{3}\:} {and}\:{v}={y}^{\mathrm{3}\:} .\:{then}\:{we}\:{have}\:{u}+{v}=\mathrm{4}\: \\ $$$${and}\:{u}^{\mathrm{3}} ×{v}^{\mathrm{3}} =\mathrm{1}.\:{so}\:{i}\:{think}\:{your}\:{can}\:{solving} \\ $$$${this}\:{problem}. \\ $$
Commented by benjo last updated on 25/Dec/19
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 25/Dec/19
$${y}=\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{4} \\ $$$${x}^{\mathrm{6}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}=\sqrt[{\mathrm{3}}]{{t}} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{3}}};\:{x}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x}_{\mathrm{1}} ;\:{x}_{\mathrm{5},\:\mathrm{6}} =\left(−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){x}_{\mathrm{2}} \\ $$$$\mathrm{and}\:{y}_{{n}} =\frac{\mathrm{1}}{{x}_{{n}} } \\ $$
Commented by benjo last updated on 25/Dec/19
$$\mathrm{it}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{6}\:\mathrm{roots} \\ $$$$\mathrm{sir}? \\ $$
Commented by MJS last updated on 25/Dec/19
$$\mathrm{yes} \\ $$