Question Number 153245 by liberty last updated on 06/Sep/21
$$\:\:\begin{cases}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}=\mathrm{30}}\\{{y}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{10}}\end{cases} \\ $$$$\:\left({x},{y}\right)=? \\ $$
Answered by Rasheed.Sindhi last updated on 06/Sep/21
$$\:\:\begin{cases}{{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}=\mathrm{30}….\left({i}\right)}\\{{y}^{\mathrm{3}} −\mathrm{3}{xy}^{\mathrm{2}} =\mathrm{10}….\left({ii}\right)}\end{cases}\:\:\:\:;\left({x},{y}\right)=? \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}+\mathrm{3}{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} =\mathrm{20} \\ $$$$\left({x}−{y}\right)^{\mathrm{3}} =\mathrm{20} \\ $$$${x}−{y}=\sqrt[{\mathrm{3}}]{\mathrm{20}} \\ $$$${y}={x}−\sqrt[{\mathrm{3}}]{\mathrm{20}} \\ $$$$\left({i}\right)\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} \left({x}−\sqrt[{\mathrm{3}}]{\mathrm{20}}\right)=\mathrm{30} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{20}}\:{x}^{\mathrm{2}} −\mathrm{30}=\mathrm{0} \\ $$$$…… \\ $$$$\:\:\:…… \\ $$$$ \\ $$
Commented by amin96 last updated on 06/Sep/21
$${x}=\sqrt[{\mathrm{3}}]{\mathrm{20}}+{y}\:\:\Rightarrow\:\:{y}^{\mathrm{3}} −\mathrm{3}\left(\sqrt[{\mathrm{3}}]{\mathrm{20}}+{y}\right){y}^{\mathrm{2}} =\mathrm{10} \\ $$$${y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{20}}−{y}^{\mathrm{3}} =\mathrm{10}\:\:\Rightarrow\:\:\mathrm{3}{y}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{20}}=−\mathrm{10}\:\: \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{10}{i}^{\mathrm{2}} }{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{20}}}\:\:\:{y}^{\mathrm{2}} =\frac{\mathrm{10}\sqrt[{\mathrm{3}}]{\mathrm{20}^{\mathrm{2}} }{i}^{\mathrm{2}} }{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{20}}\centerdot\sqrt[{\mathrm{3}}]{\mathrm{20}^{\mathrm{2}} }}\:\:\Rightarrow\:\:{y}^{\mathrm{2}} =\frac{\mathrm{10}{i}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{400}}}{\mathrm{60}} \\ $$$${y}^{\mathrm{2}} =\frac{{i}^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\mathrm{400}}}{\mathrm{6}}\:\:\Rightarrow{y}={i}\frac{\sqrt[{\mathrm{3}}]{\mathrm{20}}}{\:\sqrt{\mathrm{6}}}\:\:\ldots \\ $$
Commented by liberty last updated on 06/Sep/21
$${i}\:{got}\:{aproximation}\:{value} \\ $$
Answered by mr W last updated on 06/Sep/21
$${let}\:{u}={x}+{y},\:{v}={xy} \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}+\mathrm{3}{xy}^{\mathrm{2}} −{y}^{\mathrm{3}} =\mathrm{20} \\ $$$$\left({x}−{y}\right)^{\mathrm{3}} =\mathrm{20} \\ $$$${x}−{y}=\sqrt[{\mathrm{3}}]{\mathrm{20}} \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{50}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{50}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} −\mathrm{4}{v}=\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{50}}\:\:\:…\left({I}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{40} \\ $$$$\left({x}+{y}\right)^{\mathrm{3}} −\mathrm{6}{xy}\left({x}+{y}\right)=\mathrm{40} \\ $$$$\Rightarrow{u}^{\mathrm{3}} −\mathrm{6}{uv}=\mathrm{40}\:\:\:…\left({II}\right) \\ $$$$\mathrm{6}{u}×\left({I}\right)−\mathrm{4}\left({II}\right): \\ $$$${u}^{\mathrm{3}} −\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{50}}{u}+\mathrm{80}=\mathrm{0} \\ $$$${u}={x}+{y}=\sqrt[{\mathrm{3}}]{\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{40}}−\sqrt[{\mathrm{3}}]{\mathrm{20}\sqrt{\mathrm{3}}+\mathrm{40}} \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{20}\sqrt{\mathrm{3}}−\mathrm{40}}−\sqrt[{\mathrm{3}}]{\mathrm{20}\sqrt{\mathrm{3}}+\mathrm{40}}+\sqrt[{\mathrm{3}}]{\mathrm{20}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}−\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}+\mathrm{2}}+\mathrm{1}\right)\sqrt[{\mathrm{3}}]{\mathrm{20}}}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}−\mathrm{2}}−\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{3}}+\mathrm{2}}−\mathrm{1}\right)\sqrt[{\mathrm{3}}]{\mathrm{20}}}{\mathrm{2}} \\ $$
Commented by liberty last updated on 06/Sep/21
$${thank}\:{you} \\ $$