Transform the expression $(2a-1)(bc-1)(de-1)- (2ace-1)(b-1)(d-1)$ to can we transform the expression $$(2a-1)(bc-1)(de-1)- (2ace-1)(b-1)(d-1)$$ to some products? like $x\cdot y$? if it is impossible, is there a way to prove it?
 A: Lemma 1
If $p=p_1 \,p_2$ is a multivarate polynomial with variables $x_1,x_2,x_3,\ldots$  that is the product of two polynomials $p_1$ and $p_2$ then :


*

*the set of variable of $p_1$ and the set of variable of $p_2$ are disjoint

*no variable has  an exponent higher than $1$ in $p_1$ or $p_2$


is equivalent to


*

*no variable has an exponent higher than $1$ in $p$


Lemma 2
If $p_1(x_1,\ldots,x_i)$ and $p_2(y_1,\ldots,y_j)$ are polynomials with two different sets of variables ${x_1,\ldots,x_n}$ and ${y_1,\ldots,y_j}$ then 
the coefficient of $c_{1,\ldots,i,1,\ldots,j}$ of $x_1^{e_1}\cdots,x_i^{e_i} y_1^{f_1}\cdots y_j^{f_j}$ in $p=p_1\;p_2$ is
$a_{1,\ldots,i}\;b_{1,\ldots,j}$, where $a_{1,\ldots,i}$ is the coefficient of $x_1^{e_1}\cdots,x_i^{e_i}$ in $p_1$
and  $b_{1,\ldots,j}$ is the coefficient of $y_1^{f_1}\cdots y_j^{f_j}$ in $p_2$.
Let us assume that 
$$p=-b\,c\,d\,e+2\,a\,c\,d\,e-2\,a\,d\,e+d\,e+2\,a\,b\,c\,e-2\,a\,c\,e+
 b\,d-d-2\,a\,b\,c+b\,c-b+2\,a$$ 
can be factored in 
two polynomials $p_1\,p_2$. $p$ is linear in all variables so from Lemma 1 we conlude that $p_1$ and $p_2$ are polynomials over 
a mutual disjoint set of variables. $p$ contains the summand $-b$ so from Lemma 2 we conclude that one factor contains 
the summand $\pm b$ and the other contains the summand $\mp 1$. We can assume the constant in $p_2$ is $+1$. Otherwise we multiply both $p_1$ and $p_2$ by $-1$.
 The summand that contains $b$ is called $p_1$ and the 
summand that contains $1$ is $p_2$.
For the same reasons we conclude that one factor contains the summand $\pm d$ and the other contains the summand $\mp 1$.
If both $p_1$ and $p_2$ contain a constant summand then the product $p$ must contain a constaboth nt summand but this is not the case.
So only $p_2$ contains a constant summand and this is $1$ or $-1$. $b$ and $d$ are therefore variables of $p_1$ and not of $p_2$. 
$2a$ is a summand of $p$ so $2a$ is a summand of $p_1$.
$de$ is a summand of $p$ so either $de$ is a summand of $p_1$ or $-d$ is a summand of $p_2$. The latter is not possible becaus 
then $b\,e$ and $-2a\,e$ must be summands of the product $p$ which is not the case. So $d\,e$ is a summand of $p_1$ 
and therefore $e$ not a variable of $p_2$. A similar argument shows that $b\,c$ is a summand of $p_1$ and 
therefor $c$ is not a variable in $p_2$. So $p_2$ does not contain any of the variables $a,b,c,d,e$ and so it is identical to $1$.
Factoring multivariate polynomials can be done for small exponents 
 as noted in Factorize the polynomial $x^3+y^3+z^3-3xyz$
A: I don't think so. In my opinion the best way would be just to simplify it.
$$(2a-1)(bc-1)(de-1)- (2ace-1)(b-1)(d-1)$$
$$=(2abc-2a-bc+1)(de-1)-(2abce-2ace-b+1)(d-1)$$
$$=(2abcde-2abc-2ade+2a-bcde+bc+de-1)-(2abcde-2abce-2acde+2ace-bd+b+d-1)$$
$$=-2abc-2ade+2a-bcde+bc+de+2abce+2acde-2ace+bd-b-d$$
$$=2abce+2acde-bcde-2abc-2ade-2ace+bc+de+bd+2a-b-d$$
$$=ce(2ab+2ad-bd)-2a(bc+de+ce)+b(c+d+e)+(2a-b-d).$$
Here, find the same terms before expanding these. You'll find the one.
