1. Let $\triangle A B C$ and $\triangle A^{\prime} B^{\prime} C^{\prime}$ have side lengths $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively, and their areas be $S$ and $S^{\prime}$. If for all values of $x$, $a x^{2}+b x+c=3\left(a^{\prime} x^{2}+b^{\prime} x+c^{\prime}\right)$ always holds, then $\frac{S}{S^{\prime}}$ $=$
II. 1. From the given, $\left(a-3 a^{\prime}\right) x^{2}+\left(b-3 b^{\prime}\right) x+(c$ $\left.-3 c^{\prime}\right)=0$ is an identity, hence $\left(a-3 a^{\prime}\right)=0,\left(b-3 b^{\prime}\right)=0$, $\left(c-3 c^{\prime}\right)=0$, which gives $\frac{a}{a^{\prime}}=\frac{b}{b^{\prime}}=\frac{c}{c^{\prime}}=3$,
$\therefore \triangle A B C \sim \triangle A^{\prime} B^{\prime} C^{\prime}$, then $\frac{S}{S^{\prime}}=\left(\frac{a}{a^{\prime}}\right)^{2}=9$.