Brazilian Math Olympiad

a. First notice that, since $r$ is a root, then $r^3+r^2-4r+1=0⟺r^2+r-3=1-\frac{1}{r}$. So we need to prove that $$\begin{array}{rl}& {\left(1-\frac{1}{r}\right)}^3+{\left(1-\frac{1}{r}\right)}^2-4\left(1-\frac{1}{r}\right)+1=0\\ ⟺& 1-\frac{3}{r}+\frac{3}{r^2}-\frac{1}{r^3}+1-\frac{2}{r}+\frac{1}{r^2}-4+\frac{4}{r}+1=0\\ ⟺& -1-\frac{1}{r}+\frac{4}{r^2}-\frac{1}{r^3}=0\\ ⟺& r^3+r^2-4r+1=0,\end{array}$$ which is true.

b. Iterating $1-\frac{1}{r}$, we get $1-\frac{1}{1-\frac{1}{r}}=\frac{1}{1-r}$. It's not hard to see that $r,1-\frac{1}{r}=\frac{r-1}{r}$ and $\frac{1}{1-r}$ are all distinct. In fact, if $r=1-\frac{1}{r}$ then $r^2-r+1=0$, and $r^3=-1$, so $r^2-4r=0$, which is not true. So there are two possible ways of computing $\frac{\alpha }{\beta }+\frac{\beta }{\gamma }+\frac{\gamma }{\alpha }$: $(\alpha ,\beta ,\gamma )=(r,\frac{r-1}{r},\frac{1}{1-r})$: $$\begin{array}{rl}\frac{\alpha }{\beta }+\frac{\beta }{\gamma }+\frac{\gamma }{\alpha }& =\frac{r^2}{r-1}-\frac{(r-1{)}^2}{r}-\frac{1}{r(r-1)}\\ & =\frac{r^3-(r-1{)}^3-1}{r(r-1)}=\frac{3r(r-1)}{r(r-1)}=3\end{array}$$

$(\alpha ,\beta ,\gamma )=(r,\frac{1}{1-r},\frac{r-1}{r})$: $$\begin{array}{rl}\frac{\alpha }{\beta }+\frac{\beta }{\gamma }+\frac{\gamma }{\alpha }& =-r(r-1)-\frac{r}{(r-1{)}^2}+\frac{r-1}{r^2}\\ & =\frac{-r^6+3r^5-3r^4+r^3-3r^2+3r-1}{r^4-2r^3+r^2}\end{array}$$

Since $-r^6+3r^5-3r^4+r^3-3r^2+3r-1=(r^3+r^2-4r+1)\cdot (-r^3+4r^2-11r+29)-80r^2+130r-30=-80r^2+130r-30$ and $r^4-2r^3+r^2=(r^3+r^2-4r-1)\cdot (r-3)+8r^2-13r+3=8r^2-13r+3$. So $$\frac{\alpha }{\beta }+\frac{\beta }{\gamma }+\frac{\gamma }{\alpha }=\frac{-80r^2+130r-30}{8r^2-13r+3}=-10.$$

Another way to solve this problem is realizing that if $(\alpha ,\beta ,\gamma )=(r,\frac{r-1}{r},\frac{1}{1-r})$ then the other sum is actually $\frac{\beta }{\alpha }+\frac{\gamma }{\beta }+\frac{\alpha }{\gamma }$. By Vieta's formula, ${\sigma }_1=\alpha +\beta +\gamma =-1$, ${\sigma }_2=\alpha \beta +\beta \gamma +\gamma \alpha =-4$ and ${\sigma }_3=\alpha \beta \gamma =-1$. $$\begin{array}{rl}\frac{\alpha }{\beta }+\frac{\beta }{\gamma }+\frac{\gamma }{\alpha }+\frac{\beta }{\alpha }+\frac{\gamma }{\beta }+\frac{\alpha }{\gamma }& =\frac{{\alpha }^2\beta +{\alpha }^2\gamma +{\beta }^2\alpha +{\beta }^2\gamma +{\gamma }^2\alpha +{\gamma }^2\beta }{\alpha \beta \gamma }\\ & =\frac{{\sigma }_1{\sigma }_2-3{\sigma }_3}{{\sigma }_3}=\frac{(-1)\cdot (-4)-3(-1)}{-1}=-7,\end{array}$$ so $$ \frac{\beta}{\alpha} + \frac{\gamma}{\beta} + \frac{\alpha}{\gamma} = -7 - 3 = -10.