Example 5 Calculate:
$$
\frac{\left(2^{4}+2^{2}+1\right)\left(4^{4}+4^{2}+1\right) \cdots\left(2012^{4}+2012^{2}+1\right)}{\left(3^{4}+3^{2}+1\right)\left(5^{4}+5^{2}+1\right) \cdots\left(2013^{4}+2013^{2}+1\right)} \text {. }
$$
Notice that,
$$
\begin{aligned}
& \frac{(2 n)^{4}+(2 n)^{2}+1}{(2 n+1)^{4}+(2 n+1)^{2}+1} \\
= & \frac{\left[(2 n)^{2}+1\right]^{2}-(2 n)^{2}}{\left[(2 n+1)^{2}+1\right]^{2}-(2 n+1)^{2}} \\
= & \frac{\left(4 n^{2}+2 n+1\right)\left[(2 n)^{2}-2 n+1\right]}{\left[(2 n+1)^{2}+(2 n+1)+1\right]\left(4 n^{2}+2 n+1\right)} \\
= & \frac{(2 n)^{2}-2 n+1}{(2 n+1)^{2}+(2 n+1)+1} \\
= & \frac{(2 n-1)^{2}+(2 n-1)+1}{(2 n+1)^{2}+(2 n+1)+1} .
\end{aligned}
$$

Substituting \( n=1,2, \cdots, 1006 \) into the general term and multiplying them together, we get
the original expression
$$
\begin{array}{l}
=\frac{1^{2}+1+1}{3^{2}+3+1} \times \frac{3^{2}+3+1}{5^{2}+5+1} \times \cdots \times \frac{2011^{2}+2011+1}{2012^{2}+2012+1} \\
=\frac{3}{2013^{2}+2013+1}=\frac{3}{4054183} .
\end{array}
$$