Đáp án: $P=0$
Giải thích các bước giải:
Ta có:
$\begin{aligned} (a+b+c)(ab+bc+ca)-abc&=a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc \\&=a^2(b + c) + bc(b + c) + a(b^2 + c^2 +2bc)\\&=a^2(b + c) + bc(b+ c) +a(b + c)^2\\&=(b + c)(a^2 + bc + ab + ac)\\&=(b + c)\left[a(a + b) + c(a + b)\right]\\&=(a + b)(b + c)(c + a) \end{aligned}$
$\rightarrow (a+b)(b+c)(c+a)=2019-2019=0$
Mà:
$\begin{split}P&=(b^2c+2019)(c^2a+2019)(a^2b+2019)\\&=(b^2c+abc)(c^2a+abc)(a^2b+abc)\\&=bc(b+a).ac(c+b).ab(a+c)\\&=(abc)^2(a+b)(b+c)(c+a)\\&=0\end{split}$
