Given $\triangle A B C$, extend the three sides by 1, 2, 3 times respectively, to get $\triangle A^{\prime} B^{\prime} C^{\prime}$. Ask how many times the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is compared to the area of $\triangle A B C$.
Theorem: Let the area of $\triangle ABC$ be $S$. Extend the sides $AB, BC, CA$ of $\triangle ABC$ such that $BB'=\lambda_{1} AB$, $CC'=\lambda_{2} BC$, $AA'=\lambda_{3} CA$, and let $\triangle A' B' C'$ be the resulting triangle with area $S'$. Then,
$$
S'=1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1}.
$$

The conclusion of the theorem is a formula for the area ratio of two triangles. The area of the new triangle is the sum of the areas of four smaller triangles. If we connect $A' B, B' C, C' A$, then as shown in the figure, connecting $A' B, B' C, C' A$ results in:

In $\triangle C' B' C$ and $\triangle C B' B$, we have

Similarly, we can obtain:
$$
\begin{array}{l}
\frac{S \wedge C' A'}{S}=\lambda_{2}+\lambda_{2} \lambda_{3}. \\
\frac{S \triangle A A' B'}{S}=\lambda_{3}+\lambda_{3} \lambda_{1}.
\end{array}
$$

Therefore,
$$
\begin{array}{l}
=\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3}+\lambda_{3} \lambda_{1} \\
\end{array}
$$

Thus,
$$
\begin{aligned}
\frac{S'}{S}= & 1+\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{1} \lambda_{2}+\lambda_{2} \lambda_{3} \\
& +\lambda_{3} \lambda_{1}.
\end{aligned}
$$

Using this theorem to solve the initial problem: $\lambda_{1}=1, \lambda_{2}=2, \lambda_{3}=3$, substituting into (*) we get
$$
\frac{S_{\triangle A' B' C'}}{S_{\triangle ABC}}=18.
$$