The distinct real numbers $x_1$, $x_2$, $x_3$ satisfying the cubic equation $\alpha x^3 + \beta x^2 -3x +1 = 0$ are in harmonic progression. What are the ranges (upper and lower bounds) of $\alpha$ and $\beta$?
Four possible values of the pair $\left(\alpha,\,\beta\right)$ are provided: $A\left(-1,\,3\right)$, $B\left(2,\,-1000\right)$, $C\left(2^{2011},\,-2^{2011}\right)$ and $\left(\sqrt 5,\,\sqrt 10\right)$. Which of the values can $\left(\alpha,\,\beta\right)$ take?
(Zero or more alternatives may be correct.)
Hint: that's the same as the reciprocals of the roots of $\,P(x)=\alpha x^3+\beta x^2 - 3x + 1 =0\,$ being in arithmetic progression. The reciprocals are the roots of $\,x^3 P(\frac{1}{x})=x^3-3x^2+\beta x + \alpha=0\,$, and they sum up to $\,3\,$ by Vieta's relations. Then one root must be $\;\cdots$