8. (3 points) Fill in the circles in the figure with $10, 15, 20, 30, 40$, and $60$ so that the product of the 3 numbers at the vertices of the three small triangles $A, B, C$ are all equal. What is the maximum equal product? $\qquad$ .
【Analysis】Let the product of the 3 numbers at the vertices of the three small triangles $A, B, C$ be $S$. The middle 3 numbers are counted twice, and let the product of these three numbers be $d$. Then, $(10 \times 15 \times 20 \times 30 \times 40 \times 60) \times d=S^{3}$, which can be rewritten as $2^{9} \times 5^{6} \times 3^{3} \times d=S^{3}$. Since $2^{9}, 5^{6}, 3^{3}$ are all cube numbers, $d$ must also be a cube number. To maximize the product $d$, it must include the factor 60, where $60=2^{2} \times 3 \times 5$. To make $d$ a cube number, it needs 1 more 2, 2 more 3s, and 2 more 5s, i.e., $2 \times 3^{2} \times 5^{2}=15 \times 30$. Therefore, $d=60 \times 15 \times 30$. After adjustment, the product of the 3 numbers at the vertices of the three small triangles $A, B, C$ is: $60 \times 20 \times 15=60 \times 30 \times 10=15 \times 30 \times 40=18000$; solve accordingly.

【Solution】According to the analysis,
Let the product of the 3 numbers at the vertices of the three small triangles $A, B, C$ be $S$, and the middle 3 numbers are counted twice, with the product of these three numbers being $d$,

Then, $(10 \times 15 \times 20 \times 30 \times 40 \times 60) \times d=S^{3}$,
$$
2^{9} \times 5^{6} \times 3^{3} \times d=S^{3},
$$

Since $2^{9}, 5^{6}, 3^{3}$ are all cube numbers, $d$ must also be a cube number.
To maximize the product $d$, it must include the factor 60, where $60=2^{2} \times 3 \times 5$. To make $d$ a cube number, it needs 1 more 2, 2 more 3s, and 2 more 5s, i.e., $2 \times 3^{2} \times 5^{2}=15 \times 30$,

Therefore, $d=60 \times 15 \times 30$,
After adjustment, the product of the 3 numbers at the vertices of the three small triangles $A, B, C$ is: $60 \times 20 \times 15=60 \times 30 \times 10=15 \times 30 \times 40=18000$;