Example 6 Given $a+b+c=5$,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=15, a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$

the value. ${ }^{[4]}$
$(2016$, National Junior High School Mathematics League (B Volume))
From formula (1), we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$

From formula (2), we know
$$
47-3 a b c=5(15-5) \Rightarrow a b c=-1 \text {. }
$$

And $a^{2}+a b+b^{2}$
$$
\begin{array}{l}
=(a+b)(a+b+c)-(a b+b c+c a) \\
=5(5-c)-5=5(4-c),
\end{array}
$$

Similarly, $b^{2}+b c+c^{2}=5(4-a)$,
$$
c^{2}+c a+a^{2}=5(4-b) \text {. }
$$

Therefore, the value of the required expression is
$$
\begin{array}{l} 
125(4-a)(4-b)(4-c) \\
= 125(64-16(a+b+c)+ \\
4(a b+b c+c a)-a b c) \\
= 125(64-16 \times 5+4 \times 5-(-1)) \\
= 125 \times 5=625 .
\end{array}
$$