\(\Delta=a^2-4\left(b+1\right)>0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-a\\x_1x_2=b+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\left(x_1+x_2\right)\\b=x_1x_2-1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1^3-x_2^3=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_2=x_1-3\\x_1^3-x_2^3=9\end{matrix}\right.\)
\(\Rightarrow x_1^3-\left(x_1-3\right)^3=9\)
\(\Leftrightarrow x_1^2-3x_1+2=0\Rightarrow\left[{}\begin{matrix}x_1=1\Rightarrow x_2=-2\\x_1=2\Rightarrow x_2=-1\end{matrix}\right.\)
Thế vào (1)\(\Rightarrow\left[{}\begin{matrix}a=1;b=-3\\a=-1;b=-3\end{matrix}\right.\)
