\(\left\{{}\begin{matrix}\left(x+y\right)\left(x^3+y^3+z^3\right)+1=18\left(x+y\right)\left(1\right)\\3\left(x+y\right)^2\left(y+z\right)\left(z+x\right)=9\left(x+y\right)+1\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)\left(x^3+y^3+z^3\right)+3\left(x+y\right)^2\left(y+z\right)\left(z+x\right)=27\left(x+y\right)\)
Với \(x+y=0\). Thay vào (1) ta được: \(1=0\) (vô lí)
Với \(x+y\ne0\). Ta có:
\(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)=27\)
\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3+3\left(x+y\right)\left(z^2+xy+yz+zx\right)=27\)
\(\Rightarrow\left(x+y\right)^3+3\left(x+y\right)\left(z^2+yz+zx\right)+z^3=27\)
\(\Rightarrow\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3=27\)
\(\Rightarrow\left(x+y+z\right)^3=27\)
\(\Rightarrow x+y+z=3\Rightarrow x+y=3-z\)
\(\Rightarrow\left(x+y\right)^2=\left(3-z\right)^2\)
\(\Rightarrow x^2+y^2+6z=z^2+9-2xy\left(đpcm\right)\)
