\(x+y+z=0\Leftrightarrow\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(x+y\right)\end{matrix}\right.\)
Nhân theo vế: \(xyz=-\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(\Rightarrow2=-\left(x+y\right)\left(y+z\right)\left(x+z\right)\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=-2\)