a) x3+4x2+x-6=0
<=> x3+3x2+x2+3x-2x-6=0
<=> x2(x+3)+x(x+3)-2(x+3)=0
<=> (x+3)(x2+x-2)=0
<=> \(\left[\begin{matrix}x+3=0\\x^2+x-2=0\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=-3\\\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=-3\\x=1\\x=-2\end{matrix}\right.\)
Vậy ...
b) x3-3x2+4=0
<=> x3-2x2-x2+4=0
<=> x2(x-2)-(x-2)(x+2)=0
<=> (x-2)(x2-x-2)=0
<=> \(\left[\begin{matrix}x-2=0\\x^2-x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=2\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy ...