Given quadratic equation $5 x^3-2 x-4=0$.
Here, $a=5, b=0, c=-2, d=-4$ and $\alpha, \beta, \gamma$ are roots of equation.
$\begin{aligned}
& \alpha+\beta+\gamma=\frac{-b}{a}=0 \\
& \alpha \beta \gamma=\frac{-d}{a}=\frac{-(-4)}{5}=\frac{4}{5} \\
& \alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)+3 \alpha \beta \gamma \\
& =(0) \cdot\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alpha\right)+3 \times- \\
& \alpha^3+\beta^3+\gamma^3=\frac{12}{5}
\end{aligned}$