<s>[INST] The cubic equation $x^3 - 4 x^2 + x + 1 =0$ 
*

*The cubic polynomial
$P(x) = x^{3} - 4x^{2} + x + 1$ has discriminant $\Delta = 169 = 13^{2}$ which tells us that the extension $\mathbb{Q}(a)/\mathbb{Q}$ is normal, where $a$ is any root of the equation $P(x) = 0$.

*Therefore, given one root $a$, one can find the other as polynomial expressions in $a$.

*For instance, in this case it is not hard to check  that the other roots are $a^{2} - 4a + 2$ and $-a^{2} + 3 a + 2$. But what if we didn't know these expressions? Is there a way to get them?

 [/INST] Since $P(x) = x^3 - 4 x^2 + x + 1$ is normal, $P(x)$ splits in $\mathbb Q(\alpha)$.
This means that the polynomial will factor into linear factors over $\mathbb Q(\alpha)$. One of the factors will be $(x - \alpha)$. The others will be $(x - (\alpha^2 - 4 \alpha + 2))$ and $(x - (-\alpha^2 + 3 \alpha + 2))$
You can calculate these polynomials quickly using factorization modulo an ideal, for example in pari/gp:
? f(x) = x^3 - 4*x^2 + x + 1
%1 = (x)->x^3-4*x^2+x+1
? factor(Mod(f(x),f(a)))
%2 = 
[            x + Mod(-a, a^3 - 4*a^2 + a + 1) 1]

[x + Mod(-a^2 + 4*a - 2, a^3 - 4*a^2 + a + 1) 1]

[ x + Mod(a^2 - 3*a - 2, a^3 - 4*a^2 + a + 1) 1]


If you were just working with pencil and paper you could perform the long division of $P(x)/(x-\alpha)$ modulo $P(\alpha)$.
This gives $x^2 + (\alpha - 4) x + (\alpha^2 - 4\alpha + 1)$ which you can now try to factorize into two linear factors.
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