6.133. Without solving the equation $3 x^{2}-5 x-2=0$, find the sum of the cubes of its roots.
## Solution.

By Vieta's theorem and the given condition, we have the system

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{5}{3} \\
x_{1} \cdot x_{2}=-\frac{2}{3}
\end{array}\right.
$$

and

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

From this, $x_{1}^{3}+x_{2}^{3}=\frac{5}{3}\left(\left(\frac{5}{3}\right)^{2}-\left(-\frac{2}{3}\right)\right)=\frac{215}{27}$.

Answer: $\frac{215}{27}$.