<s>[INST] Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$. 
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
 [/INST] Since by Vieta's formulas the sum of the roots is $-2$, there is not much to do here: given that two roots are $\alpha$ and $\alpha^2+2\alpha-4$, the third one has to be $-\alpha^2-3\alpha+2$.

The interesting question, however, is: how do we realize that there are two 
   roots $\alpha,\beta$ fulfilling $\beta=\alpha^2+2\alpha-4$ ?

Well, the discriminant of the polynomial is $361=19^2$, so all the roots are real  and the Galois group over $\mathbb{Q}$ is not $S_3$ but something simpler. By considering the depressed cubic we have
$$ \frac{27}{38\sqrt{19}}\,\underbrace{p\left(\frac{2}{3}(x\sqrt{19}-1)\right)}_{q(x)}=4x^3-3x+\frac{7}{2\sqrt{19}} $$
so by trigonometry we have that a root is given by 
$$ \zeta = -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos\Big(\underbrace{\frac{1}{3}\arccos\left(\frac{-7}{2\sqrt{19}}\right)}_{\theta}\Big) $$
and the other roots are given by
$$ -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+2\pi/3)\qquad\text{and}\qquad -\frac{2}{3}+\frac{2}{3}\sqrt{19}\cos(\theta+4\pi/3) $$
i.e. by 
$$ -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)-\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta)\qquad\text{and}\qquad -\frac{2}{3}-\frac{1}{3}\sqrt{19}\cos(\theta)+\frac{1}{\sqrt{3}}\sqrt{19}\sin(\theta) $$
which are clearly related (to each other, and to $\zeta$) via the Pythagorean theorem $\sin^2\theta+\cos^2\theta=1$.
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