\(x+y=1-z\Rightarrow x^2+y^2+2xy=z^2-2z+1\)
\(\Leftrightarrow1-z^2+2xy=z^2-2z+1\)
\(\Leftrightarrow xy=z^2-z\)
Ta có:
\(1=x^3+y^3+z^3=x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)\)
\(=\left(x+y\right)^3+z^3-3\left(z^2-z\right)\left(1-z\right)\)
\(=\left(1-z\right)^3+z^3-3\left(z^2-z\right)\left(1-z\right)\)
\(=1-3z+3z^2-z^3+z^3-3\left(2z^2-z^3-z\right)\)
\(=3z^3-3z^2+1\)
\(\Rightarrow3z^3-3z^2=0\)\(\Rightarrow\left[{}\begin{matrix}z=0\\z=1\end{matrix}\right.\)
- Với \(z=0\Rightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)
- Với \(z=1\Rightarrow xy=0\Rightarrow\left[{}\begin{matrix}x=0;y=1\\x=1;y=0\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(0;0;1\right)\) và các hoán vị
\(\Rightarrow M=1\)
