, 2012

First, for $R(x) = x^3 + px + q$ to have three distinct real roots it is necessary that $p < 0$ (because the derivative $R'(x) = 3x^2 + p$ cannot be nonnegative). Let now $p < 0$, then the equation $R'(x) = 0$ has two real roots $x_1 = -\sqrt{-p/3}$, $x_2 = \sqrt{-p/3}$. Now, the condition that $R(x)$ has three distinct real roots is equivalent to the inequality $R(x_1) \cdot R(x_2) < 0$, which can be written as
$$4p^3 + 27q^2 < 0. \quad (*)$$
Let (*) be valid; denote by $\alpha_1 < \alpha_2 < \alpha_3$ the roots of $R(x)$. Since $R(x)$ is irrational, it has no zero roots. By Vieta's formula $\alpha_1 + \alpha_2 + \alpha_3 = 0$. Hence we have two possibilities:

i) $\alpha_1 < \alpha_2 < 0 < \alpha_3$;
ii) $\alpha_1 < 0 < \alpha_2 < \alpha_3$.

Without loss of generality we can assume that case i) holds (if $x^3 + px + q$ satisfies ii), then $x^3 + px - q$ satisfies i)).

So, let $\alpha_1 < \alpha_2 < 0 < \alpha_3$, then $q = -\alpha_1\alpha_2\alpha_3 < 0$. Further,
$$|\alpha_1| + |\alpha_2| + |\alpha_3| = |\alpha_1 + \alpha_2| + \alpha_3 = 2\alpha_3.$$
Hence we need to find $R(x)$ satisfying (*), $p, q \in \mathbb{Z}$, $p < 0$, $q < 0$, and for which $2\alpha_3$ is the smallest possible.

First, note that due to $p < 0$, $q < 0$ we have $R(1) = 1 + p + q < 0$ which implies $\alpha_3 > 1$. Further, $R(2) = 8 + 2p + q$. If $R(2) \le 0$, then $\alpha_3 \ge 2$. Hence, if $\alpha_3 < 2$, then
$$8 + 2p + q > 0. \quad (**)$$
Now, it is easy to see that all $R(x)$ satisfying (*), ($**$) with $p < 0$, $q < 0$ are the following: $x^3 - 2x - 1$ and $x^3 - 3x - 1$. The first of them is not irrational since it has $-1$ as a root. The second trinomial satisfies the condition.

*Remark.* One can verify that the value of $|\alpha_1| + |\alpha_2| + |\alpha_3|$ is equal to $4 \cos 20^\circ$ for the founded polynomials.