Từ \(a+\dfrac{b}{3}=-4\Rightarrow\dfrac{b}{3}=-4-a\) (1)
Ta có: \(f\left(c\right)=c^2+ac+b=-3\)
\(\Leftrightarrow c^2+ac+b+3=0\)
\(\Leftrightarrow c^2+ac+3\left(\dfrac{b}{3}+1\right)=0\)
\(\Leftrightarrow c^2+ac+3\left(-4-a+1\right)=0\) (theo 1)
\(\Leftrightarrow c^2+ac-3a-9=0\)
\(\Leftrightarrow\left(c^2-9\right)+\left(ac-3a\right)=0\)
\(\Leftrightarrow\left(c-3\right)\left(c+3\right)+a\left(c-3\right)=0\)
\(\Leftrightarrow\left(c-3\right)\left(c+3+a\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}c-3=0\\c+3+a=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=3\\a=-6\end{matrix}\right.\)
Vậy \(c=3\)
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