Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $

Question

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

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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

$$ \begin{aligned} &\alpha^6-9\alpha^4+26 \alpha^2 -24 \\ &=( 9 \alpha^2-6 \alpha+1 ) -9(3\alpha^2-\alpha) +26 \alpha^2 -24 \\ &=9\alpha^2-6\alpha+1-27\alpha^2+9 \alpha+26 \alpha^2 -24 \\&=18 \alpha^2 + 3\alpha -23 \\& = \vdots? \\&=0 \end{aligned}$$

Answer 1

$(\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$

Um... that's not right.

$(\alpha^2-2)^3-3(\alpha^2-2)+1=$

$\alpha^6 +3(-2)\alpha^4 + 3(-2)^2\alpha^2+ (-2)^3 +$

$-3\alpha^2 + 6 + $

$1 =$

$\alpha^6- 6\alpha^4 + 9\alpha^2 -1$

And

$\alpha^6- 6\alpha^4 + 9\alpha^2 -1=$

$\alpha^6 - 3\alpha^4 + \alpha^3 -3\alpha^4 -\alpha^3 +9\alpha^2 -1=$

$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 -\alpha^3 +9\alpha^2 -1=$

$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 - \alpha^3 + 9\alpha^2 -1=$

$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha^4 + 9\alpha^2 -\alpha - \alpha^3+ \alpha -1=$

$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha(\alpha^3 - 3\alpha + 1) - \alpha^3+ 3\alpha -1=$

$\alpha^3(\alpha^3 - 3\alpha + 1) -3\alpha(\alpha^3 - 3\alpha + 1) - (\alpha^3- 3\alpha +1)=$

$(\alpha^3 - 3\alpha - 1)(\alpha^3- 3\alpha +1)$

$(\alpha^3- 3\alpha +1)*0 = 0$.

Answer 2

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 $X^3-3X+1$.

Answer 3

Expand and do long division to get: $$ (x^2-2)^3-3(x^2-2)+1 = x^6 - 6 x^4 + 9 x^2 - 1 = (x^3 - 3 x - 1) (x^3 - 3 x + 1) $$