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Question Number 94174 by i jagooll last updated on 17/May/20

$$\{\begin{array}{l}\mathrm{x}+\mathrm{y}=10\\ \sqrt[3]{\mathrm{x}}+\sqrt[3]{\mathrm{y}}=\frac{5}{2}\sqrt[6]{\mathrm{xy}}\end{array}$$ $$\mathrm{find}\mathrm{x}\mathrm{\&}\mathrm{y}??$$

Commented bybehi83417@gmail.com last updated on 17/May/20

$$\mathrm{x}={\mathrm{t}}^6,\mathrm{y}={\mathrm{s}}^6$$ $$\Rightarrow \{\begin{array}{l}{\mathrm{t}}^6+{\mathrm{s}}^6=10\\ {\mathrm{t}}^2+{\mathrm{s}}^2=\frac{5}{2}\mathrm{ts}\end{array}\Rightarrow \{\begin{array}{l}({\mathrm{t}}^2+{\mathrm{s}}^2)({\mathrm{t}}^4-{\mathrm{t}}^2{\mathrm{s}}^2+{\mathrm{s}}^4)=10\\ {\mathrm{t}}^2+{\mathrm{s}}^2=\frac{5}{2}\mathrm{ts}\end{array}$$ $$\Rightarrow \{\begin{array}{l}({\mathrm{t}}^2+{\mathrm{s}}^2)[{({\mathrm{t}}^2+{\mathrm{s}}^2)}^2-{3\mathrm{t}}^2{\mathrm{s}}^2]=10\\ {\mathrm{t}}^2+{\mathrm{s}}^2=\frac{5}{2}\mathrm{ts}\end{array}$$ $$\Rightarrow \frac{5}{2}\mathrm{ts}(\frac{25}{4}{\mathrm{t}}^2{\mathrm{s}}^2-{3\mathrm{t}}^2{\mathrm{s}}^2)=10\Rightarrow {\mathrm{t}}^3{\mathrm{s}}^3=\frac{16}{13}$$ $$\Rightarrow \{\begin{array}{l}{\mathrm{t}}^6+{\mathrm{s}}^6=10\\ {\mathrm{t}}^6.{\mathrm{s}}^6=\frac{256}{169}\end{array}\Rightarrow {\mathbf{z}}^2-10\mathbf{z}+\frac{256}{169}=0$$ $$\Rightarrow \mathbf{z}=({\mathbf{t}}^6\vee {\mathbf{s}}^6)=5\pm \sqrt{25-\frac{256}{169}}=5\pm \frac{63}{13}$$ $$\Rightarrow \{\begin{array}{l}\mathrm{x}={\mathrm{t}}^6=5\pm \frac{63}{13}=\frac{128}{13},\frac{2}{13}\\ \mathrm{y}={\mathrm{s}}^6=5\pm \frac{63}{13}=\frac{128}{13},\frac{2}{13}\end{array}$$

Commented byi jagooll last updated on 17/May/20

$$\mathrm{thank}\mathrm{you}\mathrm{both}$$

Commented byhknkrc46 last updated on 17/May/20

$$\mathrm{x}+\mathrm{y}={\left(\sqrt[3]{\mathrm{x}}\right)}^3+{\left(\sqrt[3]{\mathrm{y}}\right)}^3={(\sqrt[3]{\mathrm{x}}+\sqrt[3]{\mathrm{y}})}^3-3\sqrt[3]{\mathrm{xy}}(\sqrt[3]{\mathrm{x}}+\sqrt[3]{\mathrm{y}})$$ $$={\left(\frac{5}{2}\sqrt[6]{\mathrm{xy}}\right)}^3-3\cdot \sqrt[3]{\mathrm{xy}}\cdot \frac{5}{2}\sqrt[6]{\mathrm{xy}}$$ $$=\frac{125}{8}\sqrt{\mathrm{xy}}-\frac{15}{2}\sqrt{\mathrm{xy}}=\frac{65}{8}\sqrt{\mathrm{xy}}=10$$ $$\Rightarrow \sqrt{\mathrm{xy}}=\frac{16}{13}\Rightarrow \mathrm{xy}=\frac{256}{169}$$ $$\Rightarrow \mathrm{xy}=\mathrm{x}(10-\mathrm{x})=10\mathrm{x}-{\mathrm{x}}^2=\frac{256}{169}$$ $$\Rightarrow {\mathrm{x}}^2-10\mathrm{x}+\frac{256}{169}=0\Rightarrow \mathrm{x}=\frac{10\pm \sqrt{100-4\cdot \frac{256}{169}}}{2}$$ $$\Rightarrow \mathrm{x}=5\pm \sqrt{25-\frac{256}{169}}=5\pm \frac{63}{13}(\mathrm{x},\mathrm{y}>0)$$ $$\Rightarrow \mathrm{x}=\frac{128}{13},\mathrm{x}=\frac{2}{13}\Rightarrow \mathrm{y}=\frac{2}{13},\mathrm{y}=\frac{128}{13}$$ $$\left(1\right)\mathrm{x}=\frac{128}{13}\Rightarrow \mathrm{y}=\frac{2}{13}$$ $$\left(2\right)\mathrm{x}=\frac{2}{13}\Rightarrow \mathrm{y}=\frac{128}{13}$$ $$$$

Answered by john santu last updated on 17/May/20

$$\mathrm{x},\mathrm{y}>0\mathrm{in}\in \mathbb{R}$$ $$\sqrt[6]{\frac{{\mathrm{x}}^2}{\mathrm{xy}}}+\sqrt[6]{\frac{{\mathrm{y}}^2}{\mathrm{xy}}}=\frac{5}{2}$$ $$\sqrt[6]{\frac{\mathrm{x}}{\mathrm{y}}}+\sqrt[6]{\frac{\mathrm{y}}{\mathrm{x}}}=\frac{5}{2}$$ $$\mathrm{let}\sqrt[6]{\frac{\mathrm{x}}{\mathrm{y}}}=\mathrm{t}\Rightarrow \mathrm{t}+\frac{1}{\mathrm{t}}=\frac{5}{2}$$ $${2\mathrm{t}}^2-5\mathrm{t}+2=0\Rightarrow \{\begin{array}{l}\mathrm{t}=2\\ \mathrm{t}=\frac{1}{2}\end{array}$$ $$\left(1\right)\mathrm{t}=2\Rightarrow \mathrm{x}=64\mathrm{y};\mathrm{x}+\mathrm{y}=10$$ $$65\mathrm{y}=10\{\begin{array}{l}\mathrm{y}=\frac{2}{13}\\ \mathrm{x}=\frac{128}{13}\end{array}$$ $$\left(2\right)\mathrm{t}=\frac{1}{2}\Rightarrow \mathrm{y}=64\mathrm{x};\mathrm{x}+\mathrm{y}=10$$ $$\{\begin{array}{l}\mathrm{x}=\frac{2}{13}\\ \mathrm{y}=\frac{128}{13}\end{array}$$ $$\mathrm{solution}\{(\frac{2}{13},\frac{128}{13});(\frac{128}{13},\frac{2}{13})\}.$$

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