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

$$
x_{1}^{3}+\frac{22}{x_{2}}=x_{2}^{3}+\frac{22}{x_{1}}
$$

Find all possible values of $a$.
Answer: $a= \pm 3$.

Solution. For the equation to have roots, its discriminant must be positive, from which $a^{2}+8>0$ (this inequality is true for any $a$). 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:

$$
x_{1}^{3}-x_{2}^{3}+22 \cdot \frac{x_{1}-x_{2}}{x_{1} x_{2}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)+\frac{22\left(x_{1}-x_{2}\right)}{x_{1} x_{2}}=0
$$

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 $a^{2}+2-\frac{22}{2}=0 \Leftrightarrow a^{2}-9=0$, from which $a= \pm 3$.