\(\left\{{}\begin{matrix}-\frac{b}{2a}=\frac{3}{2}\\\frac{4ac-b^2}{4a}=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=-3a\\4ac-b^2=a\end{matrix}\right.\) \(\Rightarrow4ac-9a^2=a\Rightarrow c=\frac{9a+1}{4}\)
Mặt khác theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=3\\x_1x_2=\frac{c}{a}=\frac{9a+1}{4a}\end{matrix}\right.\)
\(x_1^3+x_2^3=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=9\)
\(\Leftrightarrow27-9\left(\frac{9a+1}{4a}\right)=9\)
\(\Leftrightarrow12a-9a-1=4a\Rightarrow a=-1\)
\(\Rightarrow b=3\) ; \(c=-2\)
\(P=6\)
