x-y-10-x-1-3-y-1-3-5-2-xy-1-6-find-x-amp-y-

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

{\begin{cases} x + y = 10 \\ \sqrt[3]{x} + \sqrt[3]{y} = \frac{5}{2} \sqrt[6]{xy} \end{cases}

find x & y ? ?

Commented by behi83417@gmail.com last updated on 17/May/20

x = t^{6}, y = s^{6}

\Rightarrow {\begin{cases} t^{6} + s^{6} = 10 \\ t^{2} + s^{2} = \frac{5}{2} ts \end{cases} \Rightarrow {\begin{cases} (t^{2} + s^{2}) (t^{4} - t^{2} s^{2} + s^{4}) = 10 \\ t^{2} + s^{2} = \frac{5}{2} ts \end{cases}

\Rightarrow {\begin{cases} (t^{2} + s^{2}) [{(t^{2} + s^{2})}^{2} - {3 t}^{2} s^{2}] = 10 \\ t^{2} + s^{2} = \frac{5}{2} ts \end{cases}

\Rightarrow \frac{5}{2} ts (\frac{25}{4} t^{2} s^{2} - {3 t}^{2} s^{2}) = 10 \Rightarrow t^{3} s^{3} = \frac{16}{13}

\Rightarrow {\begin{cases} t^{6} + s^{6} = 10 \\ t^{6} . s^{6} = \frac{256}{169} \end{cases} \Rightarrow z^{2} - 10 z + \frac{256}{169} = 0

\Rightarrow z = (t^{6} \lor s^{6}) = 5 \pm \sqrt{25 - \frac{256}{169}} = 5 \pm \frac{63}{13}

\Rightarrow {\begin{cases} x = t^{6} = 5 \pm \frac{63}{13} = \frac{128}{13}, \frac{2}{13} \\ y = s^{6} = 5 \pm \frac{63}{13} = \frac{128}{13}, \frac{2}{13} \end{cases}

Commented by i jagooll last updated on 17/May/20

thank you both

Commented by hknkrc46 last updated on 17/May/20

x + y = {(\sqrt[3]{x})}^{3} + {(\sqrt[3]{y})}^{3} = {(\sqrt[3]{x} + \sqrt[3]{y})}^{3} - 3 \sqrt[3]{xy} (\sqrt[3]{x} + \sqrt[3]{y})

= {(\frac{5}{2} \sqrt[6]{xy})}^{3} - 3 \cdot \sqrt[3]{xy} \cdot \frac{5}{2} \sqrt[6]{xy}

= \frac{125}{8} \sqrt{xy} - \frac{15}{2} \sqrt{xy} = \frac{65}{8} \sqrt{xy} = 10

\Rightarrow \sqrt{xy} = \frac{16}{13} \Rightarrow xy = \frac{256}{169}

\Rightarrow xy = x (10 - x) = 10 x - x^{2} = \frac{256}{169}

\Rightarrow x^{2} - 10 x + \frac{256}{169} = 0 \Rightarrow x = \frac{10 \pm \sqrt{100 - 4 \cdot \frac{256}{169}}}{2}

\Rightarrow x = 5 \pm \sqrt{25 - \frac{256}{169}} = 5 \pm \frac{63}{13} (x, y > 0)

\Rightarrow x = \frac{128}{13}, x = \frac{2}{13} \Rightarrow y = \frac{2}{13}, y = \frac{128}{13}

(1) x = \frac{128}{13} \Rightarrow y = \frac{2}{13}

(2) x = \frac{2}{13} \Rightarrow y = \frac{128}{13}

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

x, y > 0 in \in R

\sqrt[6]{\frac{x^{2}}{xy}} + \sqrt[6]{\frac{y^{2}}{xy}} = \frac{5}{2}

\sqrt[6]{\frac{x}{y}} + \sqrt[6]{\frac{y}{x}} = \frac{5}{2}

let \sqrt[6]{\frac{x}{y}} = t \Rightarrow t + \frac{1}{t} = \frac{5}{2}

{2 t}^{2} - 5 t + 2 = 0 \Rightarrow {\begin{cases} t = 2 \\ t = \frac{1}{2} \end{cases}

(1) t = 2 \Rightarrow x = 64 y; x + y = 10

65 y = 10 {\begin{cases} y = \frac{2}{13} \\ x = \frac{128}{13} \end{cases}

(2) t = \frac{1}{2} \Rightarrow y = 64 x; x + y = 10

{\begin{cases} x = \frac{2}{13} \\ y = \frac{128}{13} \end{cases}

solution {(\frac{2}{13}, \frac{128}{13}); (\frac{128}{13}, \frac{2}{13})} .