Do \(x_1;x_2;x_3\) là nghiệm của pt nên:
\(\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)=0\)
\(\Leftrightarrow\left(x-x_1\right)\left(x^2-\left(x_2+x_3\right)x+x_2x_3\right)=0\)
\(\Leftrightarrow x^3-\left(x_1+x_2+x_3\right)x^2+\left(x_1x_2+x_2x_3+x_1x_3\right)x-x_1x_2x_3=0\)
Mà \(x^3+ax+b=0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2+x_3=0\\x_1x_2+x_2x_3+x_1x_3=a\\x_1x_2x_3=-b\end{matrix}\right.\) \(\Rightarrow x_1+x_2+x_3=0\)