Let's simplify the following expression:

$$
32 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}-b^{2}\right)^{4}+8 a b\left(a^{2}+b^{2}\right) \sqrt{16 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}-b^{2}\right)^{4}}
$$
We can separately factor out the first two terms of the expression to be transformed, $K$. Then, there remains half of the first term, which is the square of half the multiplier of the square root expression. With this, we have recognized the square of a sum:

$$
\begin{aligned}
K & =\left[4 a b\left(a^{2}+b^{2}\right)\right]^{2}+\left[16 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}-b^{2}\right)^{4}\right]+ \\
& +2 \cdot 4 a b\left(a^{2}+b^{2}\right) \sqrt{16 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}-b^{2}\right)^{4}}= \\
& =\left[4 a b\left(a^{2}+b^{2}\right)+\sqrt{16 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}-b^{2}\right)^{4}}\right]^{2}
\end{aligned}
$$

The expression under the square root, $G$, can thus be transformed as follows:

$$
\begin{aligned}
G & =16 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{4}+b^{4}-2 a^{2} b^{2}\right)^{2}= \\
& =4 \cdot 4 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left[\left(a^{2}+b^{2}\right)^{2}-4 a^{2} b^{2}\right]^{2}= \\
& =2 \cdot 4 a^{2} b^{2}\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}+b^{2}\right)^{4}+\left(4 a^{2} b^{2}\right)^{2}= \\
& =\left[\left(a^{2}+b^{2}\right)^{2}+4 a^{2} b^{2}\right]^{2}
\end{aligned}
$$

Since the expression in the square brackets is non-negative, it gives the value of the square root expression. Substituting this into (1), we can recognize another complete square, and finally,

$$
\begin{aligned}
K & =\left[4 a b\left(a^{2}+b^{2}\right)+\left(a^{2}+b^{2}\right)^{2}+4 a^{2} b^{2}\right]^{2}= \\
& =\left[\left(a^{2}+b^{2}+2 a b\right)^{2}\right]^{2}=(a+b)^{8}
\end{aligned}
$$