Factorize the following expression:

$$
\left[(a-c)^{2}+(b-d)^{2}\right] \cdot\left(a^{2}+b^{2}\right)-(a d-b c)^{2}
$$
Let the expression to be transformed be denoted by $K$. By expanding the squares of the binomials and partially multiplying:

$$
\begin{gathered}
K=\left[a^{2}+b^{2}+c^{2}+d^{2}-2(a c+b d)\right]\left(a^{2}+b^{2}\right)-a^{2} d^{2}-b^{2} c^{2}+2 a b c d= \\
=\left(a^{2}+b^{2}\right)^{2}+\left(c^{2}+d^{2}\right)\left(a^{2}+b^{2}\right)-2(a c+b d)\left(a^{2}+b^{2}\right)- \\
-a^{2} d^{2}-b^{2} c^{2}+2 a b c d
\end{gathered}
$$

Expanding the second product and simplifying, two terms remain from the product, which together with the last term form a perfect square: $a^{2} c^{2}+b^{2} d^{2}+2 a b c d=(a c+b d)^{2}$. Thus, $K$ is also a perfect square:

$$
K=\left(a^{2}+b^{2}\right)^{2}+(a b+b c)^{2}-2(a c+b d)\left(a^{2}+b^{2}\right)=\left[a^{2}+b^{2}-(a d+b c)\right]^{2}
$$

In this way, we have transformed $K$ into the product of two (equal) factors.