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

the value.
Given $a+b+c=5, a^{2}+b^{2}+c^{2}=15$, we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c \\
=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) .
\end{array}
$$

Then $47-3 a b c=5(15-5)=50$
$$
\Rightarrow a b c=-1 \text {. }
$$

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,
$$
\begin{array}{l}
b^{2}+b c+c^{2}=5(4-a), \\
c^{2}+c a+a^{2}=5(4-b) . \\
\text { Therefore }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \\
=125(4-a)(4-b)(4-c) \\
=125[64-16 \times 5+4 \times 5-(-1)] \\
=625 .
\end{array}
$$