1. Given real numbers $a b c \neq 0$, and the three quadratic equations $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ have a common root. Then $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=(\quad)$.
(A) 3
(B) 2
(C) 1
(D) 0
$-1 . A$.
Let the common root of the three quadratic equations be $x_{0}$. Substituting $x_{0}$ into the three equations and adding them yields
$$
(a+b+c)\left(x_{0}^{2}+x_{0}+1\right)=0 .
$$

Since $x_{0}^{2}+x_{0}+1=\left(x_{0}+\frac{1}{2}\right)^{2}+\frac{3}{4} \geqslant \frac{3}{4}>0$.
Therefore, $a+b+c=0$.
$$
\begin{array}{l}
\text { Hence } \frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b} \\
=\frac{-(b+c) a}{b c}+\frac{-(c+a) b}{c a}+\frac{-(a+b) c}{a b} \\
=-\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)=3 .
\end{array}
$$