Problem: Consider the polynomial $f(x) = x^3 + x^2 – 4x + 1$
a) Show that if $r$ is a root of $f$, then $r^2 + r -3$ is also a root of $f$.
b) Let, $\alpha, \beta, \gamma$ be the three roots of $f$. Determine all possible values of $\frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha}$
For the first part of this problem, I just plugged in the given expression and factorised, and it gave me $f(r)$ in the factorised expression so that was done, but it was extremely lengthy.
For the second part, I actually don't have much of a clue (the best I could do was obtain $\frac{\alpha^2 \beta + \beta^2 \gamma + \gamma^2 \alpha}{\alpha \beta \gamma}$, and hints regarding part (b) is my main problem.
Best Answer
Here is another way, using (a) to solve (b):
a) We have $x^2+x-4+\dfrac1x=0 \implies x^2+x-3=1-\dfrac1x$. Hence $$f(r^2+r-3)=f\left(1-\frac1r\right) = \dfrac{(r-1)^3+r(r-1)^2-4r^2(r-1)+r^3}{r^3}\\=-\frac{f(r)}{r^3}=0.$$
b) Given (a), the roots must be of form $p=r^2+r-3, q=p^2+p-3, r= q^2+q-3$. Then, $\frac{p}q+\frac{q}r+\frac{r}p=$ $$\frac{p^2r+q^2p+r^2q}{pqr}=-((q-p+3)r+(r-q+3)p+(p-r+3)q)\\=-3\sum_{cyc} p=3$$
Alternately, changing the order non-cyclically, $\frac{q}p+\frac{p}r+\frac{r}q=$ $$\frac{q^2r+p^2q+r^2p}{pqr}=-((r-q+3)r+(q-p+3)q+(p-r+3)p)\\=-\sum_{cyc} p^2+\sum_{cyc} pq-3\sum_{cyc} p = -10$$