Problem 1618

Let us call $\alpha \le \beta \le \gamma$ the roots of the polynomial. From the condition of being in arithmetic progression we have that there exists a nonnegative real number $\delta$ such that $\alpha = \beta - \delta$ and $\gamma = \beta + \delta$. On the other hand, using Cardano–Viète's formulas, it follows that
$$\alpha\beta\gamma = -a = \alpha + \beta + \gamma,$$
or equivalently
$$\beta(\beta^2 - \delta^2) = 3\beta.$$
Since the roots are nonzero, we finally arrive at
$$(\beta - \delta)(\beta + \delta) = \beta^2 - \delta^2 = 3.$$
We have to analyze the three possible cases:

a. If $\beta - \delta = 7/4$, then $\beta + \delta = 12/7$. From here we get that $\beta = 97/56$ and $\delta = -1/56$, which contradicts the hypothesis that $\delta \ge 0$ (alternatively, it is not possible that $\beta + \delta < \beta - \delta$).

b. If $\beta = 7/4$, then $\delta^2 = 1/4$. Therefore, the roots of the polynomial are $(3/2, 7/4, 2)$ and we obtain that
$$p(x) = (x - 3/2)(x - 7/4)(x - 2) = x^3 - \frac{21}{4}x^2 + \frac{73}{8}x - \frac{21}{4}.$$

c. If $\beta + \delta = 7/4$, then $\beta - \delta = 12/7$. From here we get that $\beta = 97/56$ and $\delta = 1/56$. Therefore, the roots of the polynomial are $(12/7, 97/56, 7/4)$ and we obtain that
$$p(x) = (x - 12/7)(x - 97/56)(x - 7/4) = x^3 - \frac{291}{56}x^2 + \frac{14113}{1568}x - \frac{291}{56}.$$