\(\left\{ {\begin{array}{*{20}{c}}{{x^2} + x\sqrt[3]{{x{y^2}}} = 80,}\\{{y^2} + y\sqrt[3]{{{x^2}y}} = 5\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}\left( {1 + \sqrt[3]{{\dfrac{{{y^2}}}{{{x^2}}}}}} \right) = 80,}\\{{y^2}\left( {1 + \sqrt[3]{{\dfrac{{{x^2}}}{{{y^2}}}}}} \right) = 5.\,\,\,}\end{array}} \right.\)
Разделим первое уравнение на второе:
\(\dfrac{{{x^2}}}{{{y^2}}} \cdot \dfrac{{1 + \sqrt[3]{{\dfrac{{{y^2}}}{{{x^2}}}}}}}{{1 + \sqrt[3]{{\dfrac{{{x^2}}}{{{y^2}}}}}}} = 16.\)
Пусть \(\dfrac{{{x^2}}}{{{y^2}}} = t\), где \(t \ge 0.\) Тогда:
\(t \cdot \dfrac{{1 + \dfrac{1}{{\sqrt[3]{t}}}}}{{1 + \sqrt[3]{t}}} = 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\dfrac{{t\left( {\sqrt[3]{t} + 1} \right)}}{{\sqrt[3]{t}\left( {1 + \sqrt[3]{t}} \right)}} = 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt[3]{{{t^2}}} = 16\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -64 < 0,}\\{t = 64.\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\)
\(\left\{ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}}}{{{y^2}}} = 64,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{y^2}\left( {1 + \sqrt[3]{{\dfrac{{{x^2}}}{{{y^2}}}}}} \right) = 5}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}}}{{{y^2}}} = 64,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{{y^2}\left( {1 + \sqrt[3]{{64}}} \right) = 5}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\dfrac{{{x^2}}}{{{y^2}}} = 64,}\\{{y^2} = 1\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 8,}\\{y = 1,}\end{array}\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -8,}\\{y = 1,\,\,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = 8,\,\,\,}\\{y = -1,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -8,}\\{y = -1.}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {8;1} \right),\,\,\,\left( {-8;1} \right),\,\,\,\left( {8;-1} \right),\,\,\,\left( {-8;-1} \right).\)