Questão 5 - 1ª Fase - IME 2025

Pelas relações de Girard, tem-se:

(I) $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=-6$

(II) $\mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=-6$

(III) $\mathrm{\alpha \beta \gamma }=3$

Dado (I), note que:

$\mathrm{\alpha }+\mathrm{\beta }=-6-\mathrm{\gamma }; \mathrm{\beta }+\mathrm{\gamma }=-6-\mathrm{\alpha }; \mathrm{\gamma }+\mathrm{\alpha }=-6-\mathrm{\beta }$

Desse modo:

$\mathrm{P}=\left(\mathrm{\alpha }+\mathrm{\beta }\right)\left(\mathrm{\beta }+\mathrm{\gamma }\right)\left(\mathrm{\gamma }+\mathrm{\alpha }\right)=\left(-6-\mathrm{\gamma }\right)\left(-6-\mathrm{\alpha }\right)\left(-6-\mathrm{\beta }\right)=-\left(6+\mathrm{\gamma }\right)\left(6+\mathrm{\alpha }\right)\left(6+\mathrm{\beta }\right)$

Desenvolvendo o produto, tem-se:

$\mathrm{P}=-\left(216+36\left(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }\right)+6\left(\mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }\right)+\mathrm{\alpha \beta \gamma }\right)$

Das relações de Girard:

$$\begin{gathered} \mathrm{P}=-\left(216+36\left(-6\right)+6\left(-6\right)+3\right) \\ \boxed{\mathrm{P}=33} \end{gathered}$$