Factorize the following expression

$$
a^{2}\left(-a^{2}+b^{2}+c^{2}\right)+b^{2}\left(a^{2}-b^{2}+c^{2}\right)+c^{2}\left(a^{2}+b^{2}-c^{2}\right) .
$$
After performing the indicated multiplications, simplifying, and factoring out a negative sign, we obtain the following form:

$$
-\left(a^{4}-2 a^{2} b^{2}+b^{4}-2 a^{2} c^{2}-2 b^{2} c^{2}+c^{4}\right)
$$

This expression differs from the square of $\left(a^{2}-b^{2}-c^{2}\right)$ only in the sign of the $2 b^{2} c^{2}$ term, ignoring the initial negative sign. By adding $4 b^{2} c^{2}$ to the expression to change the sign of the problematic term, and then subtracting it, we can express the expression as the difference of two squares:

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

By factoring the difference of squares into the product of the sum and difference of the bases:

$$
\left(2 b c+a^{2}-b^{2}-c^{2}\right) \cdot\left(2 b c-a^{2}+b^{2}+c^{2}\right)
$$

Here, each factor can again be written as the difference of two squares, which, when factored again, yields the original expression as the product of four linear sums.

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