84 (1194). a) Find the sum of the cubes of the roots of the equation $2 x^{2}-$ $-5 x+1=0$.

b) Find the value of the expression $\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}$, where $x_{1}$ and $x_{2}-$ are the roots of the equation $2 x^{2}-11 x+13=0$.
Solution. a) Having made sure of the existence of the roots of the given quadratic equation ($D>0$), we will perform transformations:

$$
\begin{gathered}
x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}\right)=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)= \\
=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right) .
\end{gathered}
$$

Based on Vieta's theorem, we have $x_{1}+x_{2}=\frac{5}{2}, x_{1} x_{2}=\frac{1}{2}$. Then

$$
x_{1}^{3}+x_{2}^{3}=\left(\frac{5}{2}\right)^{3}-3 \cdot \frac{1}{2} \cdot \frac{5}{2}=11 \frac{7}{8}
$$

Remark. The sum of the cubes of the roots of the given equation can be found in another way: determine the roots of the given quadratic equation using formulas, raise them to the third power, and add them. The method described above is shorter and, most importantly, it allows students to appreciate the beauty and elegance of rational methods of solving problems.

b) Since the discriminant of the given quadratic equation is positive $(D=17)$, the equation has two real roots and we can perform transformations:

$$
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} x_{2}}=\frac{\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}}{x_{1} x_{2}}
$$

Based on Vieta's theorem, we have:

$$
\begin{gathered}
x_{1}+x_{2}=\frac{11}{2}, x_{1} x_{2}=\frac{13}{2} \\
\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}=\left(\frac{121}{4}-13\right): \frac{13}{2}=2 \frac{17}{26}
\end{gathered}
$$

Answer: a) $11 \frac{7}{8} ;$ b) $2 \frac{17}{26}$.