Помогите пожалуйста!

Решите задачу:

$1)\; \; \left \{ {{\sqrt[3]{y}-\sqrt{x}=-4} \atop {\sqrt{x}+\sqrt[3]{y}=2}} \right. \; \; \; \; \; \; \; u=\sqrt[3]{y}\; ,\; \; v=\sqrt{x} \geq 0\; \; \\\\\left \{ {{u-v=-4} \atop {u+v=2}} \right. \; \oplus \ominus \; \left \{ {{2u=-2} \atop {2v=6}} \right. \; \left \{ {{u=-1} \atop {v=3}} \right. \; \left \{ {{\sqrt[3]{y}=-1} \atop {\sqrt{x}=3}} \right. \; \left \{ {{y=-1} \atop {x=9}} \right. \\\\Otvet:\; \; (9,-1)\; .$

$2)\; \; \left \{ {{\sqrt[5]{x}-\sqrt{y}=-3} \atop {\sqrt[5]{x}+\sqrt{y}=-1}} \right. \qquad \; u=\sqrt[5]{x}\; ,\; \; v=\sqrt{y}$

$\left \{ {{u-v=-3} \atop {u+v=-1}} \right. \; \oplus \ominus \; \left \{ {{2u=-4} \atop {2v=2}} \right. \; \left \{ {{u=-2} \atop {v=1}} \right. \; \left \{ {{\sqrt[5]{x}=-2} \atop {\sqrt{y}=1}} \right. \; \left \{ {{x=-32} \atop {y=1}} \right. \\\\Otvet:\; \; (-32,1)\; .$