<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] A possible shortcut to the problem
Notice that $$\alpha^3-3 \alpha+1=0\implies \alpha^2=3-\frac1\alpha\implies\alpha^2-2=1-\frac1\alpha$$ Now to check whether this is a root, $$\begin{align}\left(1-\frac1\alpha\right)^3-3\left(1-\frac1\alpha\right)+1&=1-\frac3\alpha+\frac3{\alpha^2}-\frac1{\alpha^3}-3+\frac3\alpha+1\\&=-1+\frac3{\alpha^2}-\frac1{\alpha^3}\\&=-\frac1{\alpha^3}(\alpha^3-3\alpha+1)=0\end{align}$$ so $\alpha^2-2$ is indeed a root of the polynomial $X^3-3X+1$.
</s>