Question

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Answer 1

$\displaystyle\bf\\\left \{ {{x^{3} +y^{3} =1} \atop {x^{2} y+2xy^{2} +y^{3}=2 \right. \\\\\\\left \{ {{(x+y)\cdot(x^{2} -xy+y^{2}) =1} \atop {y\cdot(x^{2} +2xy}+y^{2})=2 \right. \\\\\\:\left \{ {{(x+y)\cdot(x^{2} -xy+y^{2}) =1} \atop {y\cdot(x +y)^{2} =2 \right. \\--------------\\\\\frac{x^{2}-xy+y^{2} }{y\cdot(x+y)} =\frac{1}{2}\\\\\\2x^{2} -2xy+2y^{2} =xy+y^{2} \\\\2x^{2} -2xy+2y^{2} -xy-y^{2}=0\\\\2x^{2} -3xy+y^{2} =0\\\\(2x^{2} -2xy)-(xy-y^{2} )=0$

$\displaystyle\bf\\2x\cdot(x-y)-y\cdot(x-y)=0\\\\(x-y)\cdot(2x-y)=0\\\\\\\left[\begin{array}{ccc}x-y=0\\2x-y=0\end{array}\right\\\\\\\left[\begin{array}{ccc}y=x\\y=2x\end{array}\right\\\\\\1)\\\\y=x\\\\x^{3} +x^{3} =1\\\\2x^{3} =1 \ \ \ \Rightarrow \ \ \ x^{3}=\frac{1}{2} \ \ \ \Rightarrow \ \ \ \\\\x_{1} = \frac{1}{\sqrt[3]{2} } \ \ \ \Rightarrow \ \ \ y_{1} =\frac{1}{\sqrt[3]{2} } \\\\2)\\\\y=2x\\\\x^{3} +(2x)^{3} =1\\\\x^{3} +8x^{3} =1$

$\displaystyle\bf\\9x^{3}=1 \ \ \ \Rightarrow \ \ \ x^{3} =\frac{1}{9} \ \ \ \Rightarrow \ \ \ \\\\\\x_{2} =\frac{1}{\sqrt[3]{9} } \ \ \ \Rightarrow \ \ \ y_{2} =2x=\frac{2}{\sqrt[3]{9} } \\\\\\1)\\\\x^{6} +y^{6} =\Big(\frac{1}{\sqrt[3]{2} } \Big)^{6} +\Big(\frac{1}{\sqrt[3]{2} } \Big)^{6} =\Big(\frac{1}{2}\Big) ^{2} +\Big(\frac{1}{2} \Big)^{2} =\\\\\\=\frac{1}{4} +\frac{1}{4} =0,5\\\\2)$

$\displaystyle\bf\\x^{6} + y^{6} =\Big(\frac{1}{\sqrt[3]{9} } \Big)^{6} +\Big(\frac{2}{\sqrt[3]{9} } \Big)^{6} =\Big(\frac{1}{9} \Big)^{2} +\frac{64}{9^{2} } =\\\\\\=\frac{1}{81} +\frac{64}{81} =\frac{65}{81} \\\\\\Otvet \ : \ 0,5 \ ; \ \frac{65}{81}$