Question

\(\left\{{}\begin{matrix}x^2+xy+y^2=3\\7\left(x^5+y^5\right)=31\left(x^3+y^3\right)\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=3\\21\left(x^5+y^5\right)=31.3\left(x^3+y^3\right)\end{matrix}\right.\)

\(\Rightarrow31\left(x^2+xy+y^2\right)\left(x^3+y^3\right)=21\left(x^5+y^5\right)\)

\(\Rightarrow10x^5+31x^4y+31x^3y^2+31x^2y^3+31xy^4+10y^5=0\)

\(\Rightarrow\left(x+y\right)\left(x+2y\right)\left(2x+y\right)\left(5x^2-2xy+5y^2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=-x\\y=-\dfrac{1}{2}x\\y=-2x\\y=x=0\end{matrix}\right.\) \(\Rightarrow...\)