$\alpha, \beta, \gamma$ are roots of $x^3+p x^2+q x+\gamma=0$, then $\alpha^3+\beta^3+\gamma^3=$ ?
$$
\alpha+\beta+\gamma=-p
$$

$$
\begin{array}{lc}
\Rightarrow & \alpha \beta+\beta \gamma+\gamma \alpha=q \\
\Rightarrow & \alpha \beta \gamma=-r \\
\Rightarrow & {[\alpha+\beta+\gamma]^3=\alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)} \\
& \quad(\alpha \beta+\beta \gamma+\gamma \alpha)+3 \alpha \beta \gamma \\
\Rightarrow \alpha^3+\beta^3+\gamma^3=(\alpha+\beta+\gamma)^3-3(-p)(q)+3(-r) \\
\Rightarrow & \alpha^3+\beta^3+\gamma^3=-p^3+3 p q-3 r
\end{array}
$$