3. The equation $x^{2}+a x+2=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,

$$
x_{1}^{3}+\frac{14}{x_{2}^{2}}=x_{2}^{3}+\frac{14}{x_{1}^{2}}
$$

Find all possible values of $a$.
Answer: $a=4$.

Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-8>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=2$. Then $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-2$.

Transform the given equation:

$$
x_{1}^{3}-x_{2}^{3}+14 \cdot \frac{x_{1}^{2}-x_{2}^{2}}{x_{1}^{2} x_{2}^{2}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+\frac{14\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)}{\left(x_{1} x_{2}\right)^{2}}=0
$$

Since the roots are distinct, $x_{1}-x_{2} \neq 0$. Dividing both sides by $x_{1}-x_{2}$ and substituting the values given above, we get $a^{2}-2-\frac{14 a}{4}=0 \Leftrightarrow 2 a^{2}-7 a-4=0$, from which $a=-0.5$ or $a=4$. Only $a=4$ satisfies the inequality $a^{2}-8>0$.