Question Number 215282 by arkanmathematics63 last updated on 02/Jan/25
$${Show}\:{that}\:{x}=\mathrm{64}\:{and}\:{y}=\mathrm{729}\: \\ $$$$\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}=\mathrm{13} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\mathrm{35} \\ $$
Commented by Ghisom last updated on 02/Jan/25
$$\mathrm{to}\:“\mathrm{show}\:\mathrm{that}\:{x}=\mathrm{64}\:\mathrm{and}\:{y}=\mathrm{729}''\:\mathrm{it}'\mathrm{s}\:\mathrm{enough} \\ $$$$\mathrm{to}\:\mathrm{insert}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{64}}+\sqrt[{\mathrm{3}}]{\mathrm{729}}=\mathrm{4}+\mathrm{9}=\mathrm{13} \\ $$$$\sqrt{\mathrm{64}}+\sqrt{\mathrm{729}}=\mathrm{8}+\mathrm{27}=\mathrm{35} \\ $$
Commented by TonyCWX08 last updated on 02/Jan/25
$${You}\:{got}\:{a}\:{point}…………………………….. \\ $$
Commented by Ghisom last updated on 02/Jan/25
$$…\mathrm{btw}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{solutions}\:\notin\mathbb{R} \\ $$
Answered by TonyCWX08 last updated on 02/Jan/25
$${Let}\:{u}={x}^{\mathrm{6}} ,\:{v}={y}^{\mathrm{6}} \\ $$$${u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{13}\Rightarrow{p}_{\mathrm{2}} =\mathrm{13} \\ $$$${u}^{\mathrm{3}} +{v}^{\mathrm{3}} =\mathrm{35}\Rightarrow{p}_{\mathrm{3}} =\mathrm{35} \\ $$$$ \\ $$$${By}\:{Newton}'{s}\:{Identity}, \\ $$$${p}_{\mathrm{1}} ={e}_{\mathrm{1}} \\ $$$$ \\ $$$${p}_{\mathrm{2}} ={e}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{e}_{\mathrm{2}} \Rightarrow{e}_{\mathrm{2}} =\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{13}}{\mathrm{2}} \\ $$$$ \\ $$$${p}_{\mathrm{3}} ={e}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{2}{e}_{\mathrm{1}} {e}_{\mathrm{2}} +\mathrm{3}{e}_{\mathrm{3}} \\ $$$$\mathrm{35}={p}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{2}\left({p}_{\mathrm{1}} \right)\left(\frac{{p}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{13}}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{39}{p}_{\mathrm{1}} −{p}_{\mathrm{1}} ^{\mathrm{3}} }{\mathrm{2}}=\mathrm{35} \\ $$$$\mathrm{39}{p}_{\mathrm{1}} −{p}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{70} \\ $$$${p}_{\mathrm{1}} =−\mathrm{7},\:\mathrm{2},\:\mathrm{5} \\ $$$$ \\ $$$${u}+{v}=−\mathrm{7} \\ $$$${u}+{v}=\mathrm{2} \\ $$$${u}+{v}=\mathrm{5} \\ $$$$ \\ $$$${Case}\:\mathrm{1}: \\ $$$${u}+{v}=−\mathrm{7}\Rightarrow{u}=−\mathrm{7}−{v} \\ $$$$\left(−\mathrm{7}−{v}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{13}\Rightarrow{No}\:{Solution}\:{in}\:{real}\:{world} \\ $$$$ \\ $$$${Case}\:\mathrm{2}: \\ $$$${u}+{v}=\mathrm{2}\Rightarrow{u}=\mathrm{2}−{v} \\ $$$$\left(\mathrm{2}−{v}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow\left({u},{v}\right)=\left(\mathrm{2},\mathrm{3}\right)\:{or}\:\left(\mathrm{3},\mathrm{2}\right) \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{64},\mathrm{729}\right)\:{or}\:\left(\mathrm{729},\mathrm{64}\right) \\ $$$$ \\ $$$${Case}\:\mathrm{3}: \\ $$$${u}+{v}=\mathrm{5}\Rightarrow{u}=\mathrm{5}−{v} \\ $$$$\left(\mathrm{5}−{v}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{13} \\ $$$$\Rightarrow\left({u},{v}\right)=\left(\frac{\mathrm{2}−\sqrt{\mathrm{22}}}{\mathrm{2}},\frac{\mathrm{2}+\sqrt{\mathrm{22}}}{\mathrm{2}}\right)\:{or}\:\left(\frac{\mathrm{2}+\sqrt{\mathrm{22}}}{\mathrm{2}},\frac{\mathrm{2}−\sqrt{\mathrm{22}}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\left({x},{y}\right)={No}\:{solution}\:{in}\:{real}\:{world} \\ $$