Example 3 Use calculus methods to factorize $\left[(a-c)^{2}+(b-d)^{2}\right]\left(a^{2}+b^{2}\right)-(a d-b c)^{2}$.
Given that the variable $c$ has the lowest degree in the known equation, we consider the known equation as a function $f(c)$ of $c$, and differentiate it with respect to $c$, yielding $f^{\prime}(c)=-2(a-c)\left(a^{2}+b^{2}\right)+2 b(a d-b c)=2 a\left(a c-b^{2}+b d-a^{2}\right)$. Integrating with respect to $c$, we get $f(c)=\left(a c-b^{2}+b d-a^{2}\right)^{2}+c_{0}$,
and $c_{0}=f(a)=(b-d)^{2}\left(a^{2}+b^{2}\right)-a^{2}(b-d)^{2}-\left(-b^{2}-b d\right)^{2}=0$.
Therefore, the original expression $=\left(a c-b^{2}+b d-a^{2}\right)^{2}$.