<s>[INST] 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.
 [/INST] 
We write the expression as polynomial in  $z^2$. This way we obtain
  \begin{align*}
P(x,y,z)&=(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2\\
&=-(x^2+y^2+z^2)\left((x+y)^2-z^2\right)\left((x-y)^2-z^2\right)-8x^2y^2z^2\\
&=-z^6+(x^2+y^2)z^4+(x^2-y^2)^2z^2-(x^2-y^2)^2(x^2+y^2)
\end{align*}

Analyzing  the   last line  and    looking for a substitution for $z^2$          which makes this polynomial vanish it's not too hard to see that
\begin{align*}
z^2=x^2+y^2
\end{align*}
does the job.   

We  obtain
  \begin{align*}
P(x,y,z)=(x^2+y^2-z^2)Q(x,y,z)
\end{align*}
  with     $Q(x,y,z)$ a homogeneous polynomial in $x,y$ and $z$.

Division of $P(x,y,z)$ by this factor gives
\begin{align*}
\frac{-z^6+(x^2+y^2)z^4+(x^2-y^2)^2z^2-(x^2-y^2)^2(x^2+y^2)}{-z^2+x^2+y^2}=z^4-(x^2-y^2)^2
\end{align*}
From the  RHS we derive the   other     irreducible   factors of $P(x,y,z)$
\begin{align*}
z^2-(x^2-y^2)\qquad\text{and}\qquad   z^2+(x^2-y^2)
\end{align*}

and conclude
  \begin{align*}
P(x,y,z)=(x^2+y^2-z^2)(y^2+z^2-x^2)(z^2+x^2-y^2)
\end{align*}

</s>