3. Let $x_{1}$ and $x_{2}$ be the roots of the equation

$$
p^{2} x^{2}+p^{3} x+1=0
$$

Determine $p$ such that the expression $x_{1}^{4}+x_{2}^{4}$ has the smallest value.
Solution. For the expression $x_{1}^{4}+x_{2}^{4}$, using Vieta's formulas, we have

$$
\begin{aligned}
x_{1}^{4}+x_{2}^{4} & =\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-2 x_{1}^{2} x_{2}^{2} \\
& =\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]^{2}-2 x_{1}^{2} x_{2}^{2} \\
& =\left(p^{2}-\frac{2}{p^{2}}\right)^{2}-\frac{2}{p^{4}}=\left(p^{4}+\frac{2}{p^{4}}\right)-4
\end{aligned}
$$

Thus, the expression $x_{1}^{4}+x_{2}^{4}$ has its minimum value if and only if the expression $p^{4}+\frac{2}{p^{4}}$ has its minimum value. For this expression, we have

$$
p^{4}+\frac{2}{p^{4}} \geq 2 \sqrt{p^{4} \cdot \frac{2}{p^{4}}}=2 \sqrt{2}
$$

from which it follows that the expression $p^{4}+\frac{2}{p^{4}}$ attains its minimum value for $p= \pm \sqrt[8]{2}$.