Question

Известно, что x^2 + xy + y^2 = 4, x^4 + x^2*y^2 + y^4 = 8. Нужно найти x^6 + x^3*y^3 + y^6. Помогите пожалуйста

Answer 1

$\left\{\begin{array}{c}x^2 + xy + y^2 = 4\\x^4 + x^2*y^2 + y^4 = 8\end{array}\right.=>\left\{\begin{array}{c}(x^2 + y^2) + xy = 4\\(x^2 + y^2)^2-(xy)^2 = 8\end{array}\right.=>\\ \left\{\begin{array}{c}(x^2 + y^2) + xy = 4\\((x^2 + y^2)-(xy))((x^2 + y^2)+(xy)) = 8\end{array}\right.=>\left\{\begin{array}{c}(x^2 + y^2) + xy = 4\\(x^2 + y^2)-(xy) = 2\end{array}\right.=>$

$\left\{\begin{array}{c}(x^2 + y^2) = 3\\(xy) = 1\end{array}\right.\\ 27=3^3=(x^2 + y^2) ^3=x^6+3x^4y^2+3x^2y^4+y^6=x^6+y^6+3(xy)^2(x^2+y^2)=x^6+y^6+3*1^2*3=x^6+y^6+9=>x^6+y^6=27-9=18=>\\ =>x^6 + x^3*y^3 + y^6=x^6+y^6+(xy)^3=18+1^3=19$

Ответ: 19

Answer 2

Решение приложено

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