Помогите!!! Решите систему уравнений:

$\left \{ {{xy+ \frac{y^3}{x}= \frac{10}{3} } \atop {xy+ \frac{x^3}{y}=5 }} \right. \\\ \left \{ {{ \frac{x^2y+y^3}{x}= \frac{10}{3} } \atop { \frac{xy^2+x^3}{y}=5 }} \right. \\\ \left \{ {{ \frac{y(x^2y+y^2)}{x}= \frac{10}{3} } \atop { \frac{x(y^2+x^2)}{y}=5 }} \right. \\\ \left \{ {{x^2+y^2=a \geq 0} \atop { \frac{x}{y} =b}} \right.$

$\left \{ {{\frac{a}{b}= \frac{10}{3} } \atop {ab=5 }} \right. \\\ b=0.3a \\\ 0.3a^2=5 \\\ a= \sqrt{ \frac{5}{0.3} } = \sqrt{ \frac{50}{3} } \\\ b=0.3 \sqrt{ \frac{50}{3} }$

$\left \{ {{x^2+y^2= \sqrt{ \frac{50}{3} } \atop { \frac{x}{y} =0.3 \sqrt{ \frac{50}{3} }} \right. \\\ \frac{x^2}{y^2} =\frac{50\cdot0.3^2}{3}=1.5 \\\ x^2=1.5y^2 \\\ 1.5y^2+y^2= \sqrt{ \frac{50}{3}} \\\ 2.5y^2= \sqrt{ \frac{50}{3}} \\\ y^2= \frac{ \sqrt{50} }{2.5 \sqrt{3} }= \sqrt{ \frac{8}{3} } \\\ y_1= \sqrt[4]{ \frac{8}{3}} \\\ y_2= -\sqrt[4]{ \frac{8}{3}}$

$x^2=1.5 \sqrt{ \frac{8}{3} } = \sqrt{ \frac{18}{3} } = \sqrt{6} \\\ x_1= \sqrt[4]{6} \\\ x_2=- \sqrt[4]{6}$

Ответ: $( \sqrt[4]{6}; \ \sqrt[4]{ \frac{8}{3}})$ ; $( -\sqrt[4]{6}; \ -\sqrt[4]{ \frac{8}{3}})$