Question Number 141521 by 7770 last updated on 19/May/21
$$\begin{cases}{\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}+\boldsymbol{{y}}}+\sqrt{\boldsymbol{{x}}−\mathrm{3}\boldsymbol{{y}}}=\mathrm{4}}\\{\mathrm{2}\boldsymbol{{x}}+\mathrm{3}\boldsymbol{{y}}=\mathrm{17}}\end{cases} \\ $$$$\boldsymbol{\mathrm{Find}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}. \\ $$
Answered by MJS_new last updated on 20/May/21
$$\mathrm{let}'\mathrm{s}\:\mathrm{try}\:{x}+{y}={a}^{\mathrm{3}} \:\wedge\:{x}−\mathrm{3}{y}={b}^{\mathrm{2}} \\ $$$$\Leftrightarrow \\ $$$${x}=\frac{\mathrm{3}{a}^{\mathrm{3}} +{b}^{\mathrm{2}} }{\mathrm{4}}\wedge{y}=\frac{{a}^{\mathrm{3}} −{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${a}+{b}=\mathrm{4}\:\Rightarrow\:{b}=\mathrm{4}−{a} \\ $$$$\mathrm{9}{a}^{\mathrm{3}} −{b}^{\mathrm{2}} =\mathrm{68}\:\Rightarrow\:{a}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{9}}{a}^{\mathrm{2}} +\frac{\mathrm{8}}{\mathrm{9}}{a}−\frac{\mathrm{28}}{\mathrm{3}}=\mathrm{0}\:\Rightarrow\:{a}=\mathrm{2} \\ $$$$\Rightarrow\:{b}=\mathrm{2} \\ $$$$\Rightarrow\:{x}=\mathrm{7}\wedge{y}=\mathrm{1} \\ $$