How to perform this manipulation? (1)        $ z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z) $
(2)        $ (x-y)(y-z)(z-x) $
How to go from STEP (1) to STEP (2). Nothing I do seems to work. I tried combining terms but that doesn't help. 
I do not want to go from step 2 to step 1. I arrived at step 1 in some question and I need to go from 1 to 2 to match my answer given in the textbook.
I see a lot of comments asking me to just expand 2 and arrive at 1. I could see that too but I am really curious to know how it is done the other way around. Added to that, if you get (1) while solving some question, you obviously have to go from 1 to 2 and not from 2 to 1.
 A: $\begin{array}\\
z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z)
&=z^2y-z^2x+xy^2-x^2y+x^2z-y^2z\\
&=z^2(y-x)+xy(y-x)+z(x^2-y^2)\\
&=z^2(y-x)+xy(y-x)+z(x+y)(x-y)\\
&=(y-x)(z^2+xy-z(x+y))\\
&=(y-x)(z^2-zx+xy-zy)\\
&=(y-x)(z(z-x)+y(x-z))\\
&=(y-x)(z-y)(z-x)\\
\end{array}
$
A: Another method :
Rewrite (1) as 
$$-yx^2  + x^2 z + x y^2 - x z^2 - y^2 z + y z^2 + \underline{ xyz - xyz}$$
Group the terms as follows :
$$(\underline{xyz - xz^2 - y^2z + yz^2} ) - ( \underline{ x^2y-x^2z-xy^2+xyz})$$
Factor out $z$ from first underlined expression and $x$ from the second :
$$ = z(xy - xz -y^2 + yz) - x(xy - xz - y^2 + yz)$$
Now it can be easily seen that the two expressions inside the brackets are identical, so factor them out :
$$=(z-x)(xy - xz -y^2 + yz)$$
$$=(z-x)(x(y-z)-y(y-z))$$
$$=(z-x)(y-z)(x-y)$$
As Desired.
So the "trick" here is to add and subtract $abc$. It can actually be observed quite easily by multiplying the brackets in $(2)$.
Hope this helps.  :-)
A: Pair the positive and negative terms to get one of the factors
\begin{eqnarray*}
z^2y+xy^2+x^2z-(x^2y+xz^2+y^2z) = z^2y -xz^2+xy^2-x^2y +x^2z-y^2z
\end{eqnarray*}
Pull off this factor
\begin{eqnarray*}
(x-y)(-z^2-xy+(x+y)z)   
\end{eqnarray*}
The content of the second bracket is easily shown to be what is required. 
A: Hint: the first expression can easily seen to be $0$ for $x=y$ and $x=z$. Considering it as a polynomial in $x\,$ of degree $2$, this means that it factors as $\lambda(x-y)(x-z)$ where the dominant coefficient $\lambda$ must match the coefficient of $x^2$ from the original expression i.e. $\lambda=z-y$.
