Giải thích các bước giải:
Ta có:
$a^2(b+c)+b^2(c+a)+c^2(a+b)+2abc=0$
$\to a^2b+a^2c+b^2c+b^2a+c^2(a+b)+2abc=0$
$\to ( a^2b+b^2a)+(a^2c+b^2c+2abc)+c^2(a+b)=0$
$\to ab(a+b)+c(a+b)^2+c^2(a+b)=0$
$\to (a+b)(ab+c(a+b)+c^2)=0$
$\to (a+b)(ab+ca+cb+c^2)=0$
$\to (a+b)(a+c)(b+c)=0$
Không mất tính tổng quát giả sử $a+b=0$
$\to a=-b$
$\to a^{2019}+b^{2019}+c^{2019}=(-b)^{2019}+b^{2019}+c^{2019}$
$\to a^{2019}+b^{2019}+c^{2019}=-b^{2019}+b^{2019}+c^{2019}$
$\to a^{2019}+b^{2019}+c^{2019}=0+c^{2019}$
$\to a^{2019}+b^{2019}+c^{2019}=c^{2019}$
$\to a^{2019}+b^{2019}+c^{2019}=(0+c)^{2019}$
$\to a^{2019}+b^{2019}+c^{2019}=(a+b+c)^{2019}$
