Factorize $(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$ I am unable to factorize this over $\mathbb{Z}:$ 
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$$
Since, this from an exercise of a book (E. J. Barbeau, polynomials) it must have a neat factorization.
I tried to guess some factors by putting $x-y=0$, $x+y=0$, $x=0$.
I also notice that the expression is symmetric with respect to $x, y, z$.
I am trying to factor it without expanding the whole thing though the process of expansion can be eased by observing that $(x+y+z)(x+y-z)(y+z-x)(z+x-y)$ is a well known expression (Heron's formula), so one already knows it is 
$$2(x^2y^2+y^2z^2+z^2x^2)-x^4-y^4-z^4$$
 But I think there should be a way to do this without expanding the thing.
So, can someone tell me the factorization (preferably with the steps) or any hints as to  how should I approach this problem.

Note that
\begin{eqnarray}
&&(x^2+y^2+z^2)(x+y+z) (x+y-z) (y+z-x) (z+x-y)-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)(-x^4+2 x^2 y^2+2 x^2 z^2-y^4+2 y^2 z^2-z^4)-8x^2y^2z^2\\
&=&(x^2+y^2+z^2)[(-x^4+2 x^2 y^2-y^4)+(2 x^2 z^2+2 y^2 z^2-z^4)]-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)[z^4-(x^2-y^2)^2+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&[(x^2+y^2-z^2)+2z^2][(z^2+x^2-y^2)(z^2-x^2+y^2)+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&[(x^2+y^2-z^2)+2z^2][(x^2-y^2+z^2)(-x^2+y^2+z^2)+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&(x^2+y^2-z^2)(x^2-y^2+z^2)(-x^2+y^2+z^2)+2z^2[(x^2-y^2+z^2)(-x^2+y^2+z^2)\\
&&+(x^2+y^2-z^2)^2+2(x^2+y^2-z^2)z^2-4 x^2 y^2]\\
&=&(x^2+y^2-z^2)(x^2-y^2+z^2)(-x^2+y^2+z^2).
\end{eqnarray}
