3. The equation $x^{2}+a x+8=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,

$$
x_{1}-\frac{64}{17 x_{2}^{3}}=x_{2}-\frac{64}{17 x_{1}^{3}}
$$

Find all possible values of $a$.
Answer: $a= \pm 12$.

Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-32>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=8$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-8$.

Transform the given equality:

$$
x_{1}-x_{2}-\frac{64}{17} \cdot \frac{x_{1}^{3}-x_{2}^{3}}{x_{1}^{3} x_{2}^{3}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)-\frac{64\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)}{17\left(x_{1} x_{2}\right)^{3}}=0
$$

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 $1-\frac{64}{17} \cdot \frac{a^{2}-8}{512}=0 \Leftrightarrow a^{2}-144=0$, from which $a= \pm 12$. Both found values of the parameter satisfy the inequality $a^{2}-32>0$.