<s>[INST] Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root 
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root

set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begin{aligned}
\alpha^6&= \alpha^3 \alpha^3
        \\&=(3\alpha-1) (3\alpha-1)
        \\&= 3\alpha(3\alpha-1)-1(3\alpha-1)
        \\&= 9\alpha^2-3\alpha-3\alpha+1
        \\&=9 \alpha^2 -6 \alpha+1
\end{aligned} $$
Now looking at $\alpha^4$
$$
\begin{aligned}
\alpha^4= \alpha^3 \alpha^1
        &=(3\alpha-1)  \alpha
        &=3 \alpha^2-\alpha
\end{aligned}
 $$
Let us go back 
$$ \begin{aligned}
&\alpha^6-9\alpha^4+26 \alpha^2 -24
\\ 
   &=( 9 \alpha^2-6 \alpha+1  )
          -9(3\alpha^2-\alpha) 
           +26 \alpha^2 -24
\\ &=9\alpha^2-6\alpha+1-27\alpha^2+9 \alpha+26 \alpha^2 -24
\\&=18 \alpha^2 + 3\alpha -23
\\& = \vdots?
\\&=0          
\end{aligned}$$
 [/INST] Expand and do long division to get:
$$
 (x^2-2)^3-3(x^2-2)+1 = x^6 - 6 x^4 + 9 x^2 - 1 = (x^3 - 3 x - 1) (x^3 - 3 x + 1)
$$
</s>