Four, (Full marks 20 points, Sub-question (1) 6 points, Sub-question (2) 14 points) Let the two real roots of $x^{2}-p x+q=0$ be $\alpha, \beta$.
(1) Find the quadratic equation whose roots are $\alpha^{3}, \beta^{3}$;
(2) If the quadratic equation whose roots are $\alpha^{3}, \beta^{3}$ is still $x^{2}-p x$ $+q=0$, find all such quadratic equations.
$\therefore$ The required equation is $x^{2}-p\left(p^{2}-3 q\right) x+q^{3}=0$.
(2) From the given and the derived results, we have $\left\{\begin{array}{l}p\left(p^{2}-3 q\right)=p, \\ q^{3}=q \Rightarrow q=0,1,-1 \text {. }\end{array}\right.$

Then
\begin{tabular}{c|c|c|c}
$p$ & $0,1,-1$ & $0,2,-2$ & 0 \\
\hline$q$ & 0 & 1 & -1
\end{tabular}

According to the seven cases of $p, q$ values, we get the following seven equations
$$
x^{2}=0, x^{2}-x=0, x^{2}+x=0, x^{2}+1=0, x^{2}-2 x
$$
$+1=0, x^{2}+2 x+1=0, x^{2}-1=0$, among which only $x^{2}+1=0$
has no real roots, and should be discarded. The remaining six equations are the required ones.
Four, (1) By Vieta's formulas, we get $\alpha+\beta=p, \alpha \beta=q$, then $\alpha^{3} \beta^{3}=$
$$
(\alpha \beta)^{3}=q^{3}, \alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)=p^{3}-3 p q .
$$