2・54 If the solutions of the equation $x^{2}+p x+q=0$ are the cubes of the solutions of the equation $x^{2}+m x+n=0$, then
(A) $p=m^{3}+3 m n$.
(B) $p=m^{3}-3 m n$.
(C) $p+q=m^{3}$.
(D) $\left(\frac{m}{n}\right)^{3}=\frac{p}{q}$.
(E) None of these.
(28th American High School Mathematics Examination, 1977)
[Solution] Let $\alpha$ and $\beta$ be the solutions of $x^{2}+m x+n=0$. Then, according to the problem, we have
$$
\left\{\begin{array}{rl}
-m = \alpha + \beta, \\
n & = \alpha \beta;
\end{array} \quad \left\{\begin{array}{rl}
-p & =\alpha^{3}+\beta^{3}, \\
q & =\alpha^{3} \beta^{3}.
\end{array}\right.\right.
$$

Since
$$
\begin{array}{c}
\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta), \\
-p=-m^{3}+3 m n.
\end{array}
$$
$$
\therefore
$$

Therefore, the answer is $(B)$.