1.50 Given $a+b+c=6, a^{2}+b^{2}+c^{2}=10, a^{3}+b^{3}+c^{3}=$ 36. Then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is
(A) $\frac{13}{18}$.
(B) $\frac{18}{13}$.
(C) $-\frac{13}{18}$.
(D) $-\frac{18}{13}$.
(Shenyang City, Liaoning Province, China "Yucai Cup" Junior High School Mathematics Competition, 1990)
$$
\begin{array}{l}
\text { [Solution] From } a+b+c=6 \text { and } a^{2}+b^{2}+c^{2}=10 \text { and } \\
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a), \\
\therefore \quad a b+b c+c a=13 . \\
\text { Also, } \quad a^{3}+b^{3}+c^{3} \\
\quad=(a+b+c)^{3}-3(a+b+c)(a b+b c+c a)+3 a b c,
\end{array}
$$
$$
\text { Therefore, } \quad a b c=18 \text {. }
$$

Thus, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}=\frac{13}{18}$.
Hence, the answer is $(A)$.