3. Find $\frac{1}{x_{1}^{5}}+\frac{1}{x_{2}^{5}}$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $x^{2}+k x+s=0$.
Solution. According to Vieta's formulas, the solutions $x_{1}$ and $x_{2}$ of the given quadratic equation satisfy the equations

$$
x_{1}+x_{2}=-k \quad \text { and } \quad x_{1} x_{2}=s
$$

Using this fact, as well as the identity

$$
x_{1}^{5}+x_{2}^{5}=\left(x_{1}+x_{2}\right)\left(x_{1}^{4}-x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}-x_{1} x_{2}^{3}+x_{2}^{4}\right)
$$

we have

$$
\begin{aligned}
\frac{1}{x_{1}^{5}}+\frac{1}{x_{2}^{5}} & =\frac{x_{1}^{5}+x_{2}^{5}}{x_{1}^{5} x_{2}^{5}}=\frac{\left(x_{1}+x_{2}\right)\left(x_{1}^{4}-x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}-x_{1} x_{2}^{3}+x_{2}^{4}\right)}{\left(x_{1} x_{2}\right)^{5}} \\
& =\frac{-k\left[\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-x_{1}^{2} x_{2}^{2}-x_{1}^{3} x_{2}-x_{1} x_{2}^{3}\right]}{s^{5}} \\
& =\frac{-k\left\{\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]^{2}-\left(x_{1} x_{2}\right)^{2}-x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}\right)\right\}}{s^{5}} \\
& =\frac{-k\left\{\left(k^{2}-2 s\right)^{2}-\left(x_{1} x_{2}\right)^{2}-s\left[\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}\right]\right\}}{s^{5}}
\end{aligned}
$$

$$
=-\frac{k\left[\left(k^{2}-2 s\right)^{2}-s^{2}-s\left(k^{2}-2 s\right)\right]}{s^{5}}=-\frac{k\left(k^{4}-5 k^{2} s+5 s^{2}\right)}{s^{5}}
$$