<=> x^3-x-6x+6=0
<=>(x^3-x)-(6x-6)=0
<=>x.(x^2-1)-6.(x-1)=0
<=>x.(x-1).(x+1)-6.(x+1)=0
<=>(x+1).(x^2-x-6)=0
<=>(x+1).(x^2+2x-3x-6)=0
<=>(x+1).[x.(x+2)-3.(x+2)]=0
<=>(x+1).(x+2).(x-3)=0
còn lại pn lm nốt nhá
x3 - 7x + 6 = 0
\(\Leftrightarrow x^3-x-6x+6=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-3x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Vậy: \(x=1;2;3\)