\(\left(a+b+c\right)\left(ab+bc+ca\right)=abc\)
\(\Rightarrow\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc\right)-abc=0\)
\(\Rightarrow a^2b+bc^2+2abc+a^2c+ac^2+b^2c+ab^2=0\)
\(\Rightarrow b\left(a+c\right)^2+ac\left(a+c\right)+b^2\left(a+c\right)=0\)
\(\Rightarrow\left(a+c\right)\left[b\left(a+c\right)+ac+b^2\right]=0\)
\(\Rightarrow\left(a+c\right)\left(a+b\right)\left(b+c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+c=0\Rightarrow a^{2019}+c^{2019}=0\\b+c=0\Rightarrow b^{2019}+c^{2019}=0\\a+b=0\Rightarrow a^{2019}+b^{2019}=0\end{matrix}\right.\)
\(\Rightarrow P=1\)
*Hằng đẳng thức cần áp dụng:
\(x^n+y^n=\left(x+y\right)\left(x^{n-1}-x^{n-2}y+...-xy^{n-2}+y^{n-1}\right)\)
nên \(x+y=0\Rightarrow x^n+y^n=0\)
