Ta có:
\(x^3+y^3+z^3=3xyz\left(gt\right)\)
\(\Rightarrow x^3+y^3+z^3-3xyz=0\)
\(\Rightarrow x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Rightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Rightarrow\left(x+y+z\right)^3-3z\left(x+y\right)\left(x+y+z\right)-3xy\left(x+y+z\right)=0\)
\(\Rightarrow\left(x+y+z\right)^3-\left(x+y+z\right)\left(3xy+3zx+3yz\right)=0\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2+2xy+2xz+2yz-3xy-3xz-3yz\right)=0\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)
\(\Rightarrow\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y+z=0\\\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
Mà \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=y\\y=z\\z=x\end{matrix}\right.\)
\(\Rightarrow x=y=z\)
Xét trường hợp x = y = z, ta có:
\(P=\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(P=\dfrac{x^3}{2x.2x.2x}\)
\(P=\dfrac{x^3}{8x^3}\)
\(P=\dfrac{1}{8}\)
Xét trường hợp x + y + z = 0, ta có:
\(\left\{{}\begin{matrix}x=-\left(y+z\right)\\y=-\left(x+z\right)\\z=-\left(y+x\right)\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{-\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Rightarrow P=-1\)
