Đặt f(x)=0
=>(x-1)(x+3)=0
=>x=1 hoặc x=-3
Theo đề, ta có:
\(\left\{{}\begin{matrix}1^3-a\cdot1^2+b\cdot1-3=0\\\left(-3\right)^3-a\cdot\left(-3\right)^2+b\cdot\left(-3\right)-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-a+b=2\\3a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4a=-6\\a-b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}\\b=a+2=\dfrac{3}{2}+2=\dfrac{7}{2}\end{matrix}\right.\)