Ta có: \(a+b+c=3\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)
\(\Rightarrow2\left(ab+bc+ca\right)=9-\left(a^2+b^2+c^2\right)=6\Rightarrow ab+bc+ca=3\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
Mà a + b + c = 3 nên a = b = c = 1
Suy ra \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)