, 2011

We have three relations:
$$\begin{aligned} a \alpha^2 - b \alpha - c &= \lambda, \\ b \alpha^2 - c \alpha - a &= \lambda, \\ c \alpha^2 - a \alpha - b &= \lambda, \end{aligned}$$
where $\lambda$ is the common value. Eliminating $\alpha^2$ from these, taking these equations pairwise, we get three relations:
$$\begin{aligned} (c a - b^2) \alpha - (b c - a^2) &= \lambda (b - a), \\ (a b - c^2) \alpha - (c a - b^2) &= \lambda (c - b), \\ (b c - a^2) - (a b - c^2) &= \lambda (a - c). \end{aligned}$$
Adding these three, we get
$$(a b + b c + c a - a^2 - b^2 - c^2)(\alpha - 1) = 0.$$
(Alternatively, multiplying above relations respectively by $b-c$, $c-a$ and $a-b$, and adding also leads to this.) Thus either $a b + b c + c a - a^2 - b^2 - c^2 = 0$ or $\alpha = 1$. In the first case
$$0 = a b + b c + c a - a^2 - b^2 - c^2 = \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2)$$
shows that $a = b = c$. If $\alpha = 1$, then we obtain
$$a - b - c = b - c - a = c - a - b,$$
and once again we obtain $a = b = c$.