Finding the value of $(bc-ad)(ac-bd)(ab-cd)$ Let $a,b,c,d$ be $4$ distinct non-zero integers such that $a+b+c+d = 0$. It is know that the number
$$M = (bc - ad)(ac - bd)(ab-cd)$$
lies strictly between $96100$ and $98000$. Determine the value of $M$.
I tried expanding the expression out, as well as using AM-GM on it, but to no avail. Any help would be appreciated. Thanks!
(Source: Singapore Mathematical Olympiad 2013, Open Section, First Round, Question 24)
 A: After the substitution of $a = -b-c-d$, we have $M = (b+c)^2(b+d)^2(c+d)^2$. 
A: We need to say a bit more after Erik's answer.  We have
$$(b+c)(b+d)(c+d)=\pm311,\,\pm312,\,\pm313\ .$$
Now the sum of the three numbers $b+c$, $b+d$, $c+d$ is even and so at least one of the numbers must be even.  Therefore the only possibility is that
$$(b+c)(b+d)(c+d)=\pm312$$
and we have
$$M=312^2\ .$$
Note: we cannot straight away rule out the possibility that the product of the three numbers is prime, because they could be $1,1,p$.
And another (possibly) interesting point.  We can of course obtain Erik's identity by plain algebra, but the form of $M$ suggests determinants, so here is an alternative method.  Using the fact that $a+b+c+d=0$ and adding columns, we have 
$$bc-ad=-\Bigl|\matrix{a&b\cr c&d\cr}\Bigr|
  =-\Bigl|\matrix{a+b&b\cr c+d&d\cr}\Bigr|=-\Bigl|\matrix{-(c+d)&b\cr c+d&d\cr}\Bigr|$$
Now taking out a factor of $-(c+d)$ and row-reducing,
$$bc-ad=(c+d)\Bigl|\matrix{1&b\cr-1&d\cr}\Bigr|
  =(c+d)\Bigl|\matrix{1&b\cr0&b+d\cr}\Bigr|
  =(c+d)(b+d)\ ,$$
and similarly for the others.
