\(x^2+y^2+z^2=1\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\le1\\y^2\le1\\z^2\le1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}-1\le x\le1\\-1\le y\le1\\-1\le z\le1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\left(x-1\right)\le0\\y^2\left(y-1\right)\le0\\z^2\left(z-1\right)\le0\end{matrix}\right.\) (1)
+ \(x^2+y^2+z^2=x^3+y^3+z^3\)
\(\Leftrightarrow x^2\left(x-1\right)+y^2\left(y-1\right)+z^2\left(z-1\right)=0\) (2)
+ Từ (1) và (2) suy ra :
\(\left\{{}\begin{matrix}x^2\left(x-1\right)=0\\y^2\left(y-1\right)=0\\z^2\left(z-1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\\left[{}\begin{matrix}y=0\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}z=0\\z=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\) trong 3 số a,b,c có 1 số bằng 1 và 2 số bằng 0
( do \(x^2+y^2+z^2=x^3+y^3+z^3=1\))
=> P = 0
