Solve the equation $x^3-3x^2-6x+8=0$ given that the roots are in AP 

Let $a-d, a, a+d$ be the roots of the given equation 

$S_1=\frac{-b}{a}\Rightarrow a-d+a+a+d=-\frac{\left(3\right)}{1}$

$$\begin{gathered} 3a=3; a=1 \\ \therefore (x-1) is a factor of x^3-3x^2-6x+8=0 \\ By synthetic division \end{gathered}$$

$$\begin{gathered} \Rightarrow x^2-2x-8=0; x^2-4x+2x-8=0 \\ x\left(x-4\right)+2\left(x-4\right)=0; \left(x-4\right)\left(x+2\right)=0 \\ x-4=0,x+2=0; x=4, x=-2 \\ The roots are -2, 1, 4 \end{gathered}$$