P(0)=2017 nên \(a\cdot0+b\cdot0+c+d=2017\)
=>d=2017
Do đó: \(P\left(x\right)=ax^3+bx^2+cx+2017\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+b+c+2017=2\\-a+b-c+2017=6\\8a+4b+2c+2017=-6033\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=-2015\\-a+b-c=-2011\\8a+4b+2c=-8050\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2013\\c=-3\end{matrix}\right.\)