3. The equation $x^{2}+a x+4=0$ has two distinct roots $x_{1}$ and $x_{2}$; in this case,

$$
x_{1}^{2}-\frac{20}{3 x_{2}^{3}}=x_{2}^{2}-\frac{20}{3 x_{1}^{3}}
$$

Find all possible values of $a$.
Answer: $a=-10$.

Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-16>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=4$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-4$.

Transform the given equation:

$$
x_{1}^{2}-x_{2}^{2}-\frac{20}{3} \cdot \frac{x_{1}^{3}-x_{2}^{3}}{x_{1}^{3} x_{2}^{3}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)-\frac{20\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)}{3\left(x_{1} x_{2}\right)^{3}}=0
$$

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 $-a-\frac{20}{3} \cdot \frac{a^{2}-4}{64}=0 \Leftrightarrow 5 a^{2}+48 a-20=0$, from which $a=-10$ or $a=0.4$. Only $a=-10$ satisfies the inequality $a^{2}-16>0$.