1.48 Given $a+b+c=0$, simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)$ $+c\left(\frac{1}{a}+\frac{1}{b}\right)+3$.

(A) 1.
(B) -1.
(C) $\pm 1$.
(D) 0.
(2nd "Molybdenum Cloud Cup" Junior High School Mathematics Invitational, 1985)
[Solution 1] We first prove the following fact:
If $\alpha+\beta+\gamma=0$, then $\alpha^{3}+\beta^{3}+\gamma^{3}=3 \alpha \beta \gamma$.
Let $\alpha, \beta, \gamma$ be the three roots of the cubic equation $x^{3}+p x+q=0$. Then by Vieta's formulas, we have $\alpha+\beta+\gamma=0, \quad \alpha \beta+\beta \gamma+\gamma \alpha=p, \quad \alpha \beta \gamma=-q$.
Substituting $\alpha, \beta, \gamma$ into the equation and adding both sides, we get
$$
\alpha^{3}+\beta^{3}+\gamma^{3}+p(\alpha+\beta+\gamma)+3 q=0 .
$$

From $\circledast$ we know
$$
\alpha^{3}+\beta^{3}+\gamma^{3}=3 \alpha \beta \gamma \text {. }
$$

Returning to our problem, note that
the original expression $=\frac{-a^{2}}{b c}+\frac{-b^{2}}{a c}+\frac{-c^{2}}{a b}+3=\frac{-\left(a^{3}+b^{3}+c^{3}\right)+3 a b c}{a b c}$.
By the previous conclusion, the above expression is 0.
Therefore, the answer is $(D)$.
[Solution 2] The original expression $=\left(\frac{a}{b}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{b}{a}\right)+\left(\frac{c}{a}+\frac{c}{b}\right)+$
$$
\begin{aligned}
& \left(\frac{a}{a}+\frac{b}{b}+\frac{c}{c}\right) \\
= & \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}=0 .
\end{aligned}
$$

Therefore, the answer is $(D)$.