2.111 If $a, b, c, d$ are the solutions of the equation $x^{4}-b x-3=0$, then the four solutions
$$
\frac{a+b+c}{d^{2}}, \quad \frac{a+b+d}{c^{2}}, \quad \frac{a+c+d}{b^{2}}, \quad \frac{b+c+d}{a^{2}}
$$

are the solutions of the equation
(A) $3 x^{4}+b x+1=0$.
(B) $3 x^{4}-b x+1=0$.
(C) $3 x^{4}+b x^{3}-1=0$.
(D) $3 x^{4}-b x^{3}-1=0$.
(E) None of the above equations.
(32nd American High School Mathematics Examination, 1981)
[Solution] Since the coefficient of the cubic term $x^{3}$ in the polynomial function $f(x)=x^{4}-b x-3$ is zero, by Vieta's formulas we have
$$
a+b+c+d=0,
$$

and $\quad \frac{a+b+c}{d^{2}}=\frac{a+b+c+d-d}{d^{2}}=\frac{-1}{d}$,
$$
\frac{a+c+d}{b^{2}}=\frac{-1}{b}, \quad \frac{a+b+d}{c^{2}}=\frac{-1}{c}, \quad \frac{b+c+d}{a^{2}}=\frac{-1}{a} \text {. }
$$

Since $a, b, c, d$ are the roots of $f(x)$,
then $\frac{-1}{d}, \frac{-1}{b}, \frac{-1}{c}, \frac{-1}{a}$ are the roots of $f\left(-\frac{1}{x}\right)=0$,
i.e., $\frac{a+b+c}{d^{2}}, \quad \frac{a+b+d}{c^{2}}, \quad \frac{a+c+d}{b^{2}}, \quad \frac{b+c+d}{a^{2}}$ are also the roots of $f\left(-\frac{1}{x}\right)=0$. This equation is
$$
\frac{1}{x^{4}}+\frac{b}{x}-3=0 \text { or } 3 x^{4}-b x^{3}-1=0 \text {. }
$$

Therefore, the answer is $(D)$.