Question

Решите уравнение, пожалуйста) ​

Answer 1

$\left \{ {{\sqrt[3]{x}+\sqrt[3]{y}=4} \atop {x+y=28}} \right. \\\\\sqrt[3]{x}=m;\sqrt[3]{y}=n\\\\\left \{ {{m+n=4} \atop {m^{3}+n^{3}=28}} \right. \\\\:\left \{ {{(m+n)(m^{2}-mn+n^{2})=28} \atop {m+n=4}} \right.\\-----------\\m^{2}-mn+n^{2}=7\\\\n=4-m\\\\m^{2} -m*(4-m) +(4-m)^{2}=7\\\\m^{2}-4m+m^{2}+16-8m+m^{2}-7=0\\\\3m^{2}-12m+9=0\\\\m^{2}-4m+3=0\\\\m_{1}=1\\\\m_{2}=3\\\\n_{1}=4-1=3\\\\n_{2}=4-3=1$

$1)\left \{ {{\sqrt[3]{x} =1} \atop {\sqrt[3]{y} =3}} \right.\\\\\left \{ {{x=1} \atop {y=27}} \right.\\\\2)\left \{ {{\sqrt[3]{x} =3} \atop {\sqrt[3]{y}=1 }} \right.\\\\\left \{ {{y=x=27} \atop {y=1}} \right. \\\\Otvet:\boxed{(27;1),(1;27)}$