Решите пожаааалуййста

Решите задачу:

$\left \{ {{x+y=72} \atop {\sqrt[3]{x}+\sqrt[3]{y}=6}} \right. \; \left \{ {{(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y})=72} \atop {\sqrt[3]{x}+\sqrt[3]{y}=6}} \right. \; \; \Rightarrow \\\\1)\quad \sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}=72:6\; \\\\ (\sqrt[3]{x^2}+\sqrt[3]{y^2})-\sqrt[3]{xy}=12$

$\sqrt[3]{x^2}+\sqrt[3]{y^2}=12+\sqrt[3]{xy}$

$2)\quad 6^2=(\sqrt[3]{x}+\sqrt[3]{y})^2=\sqrt[3]{x^2}+\sqrt[3]{y^2}+2\sqrt[3]{xy}\\\\\sqrt[3]{x^2}+\sqrt[3]{y^2}=36-2\sqrt[3]{xy}$

$3)12+\sqrt[3]{xy}=36-2\sqrt[3]{xy}$

$3\sqrt[3]{xy}=24\\\\\sqrt[3]{xy}=8\; \; \to \; \; xy=8^3\; ,\; xy=512\\\\4) \quad \left \{ {{x+y=72} \atop {xy=512}} \right. \; \left \{ {{y=72-x} \atop {x(72-x)=512}} \right. \; \left \{ {{y=72-x} \atop {x^2-72x+512=0}} \right. \; \left \{ {{y_1=64,\; y_2=8} \atop {x_1=8,\; x_2=64}} \right. \\\\Otvet:\quad (8,64)\; ;\; (64,8)\; .$