What does Heron's formula naturally deform?

Fixing three real numbers $a,b,c>0$ determines a triangle with side-lengths $a,b,c$ (if admissible). Therefore, the area of a triangle is a function in $a,b,c$. Due to the geometry of a triangle, we know that the area is a symmetric function in $a,b,c$. Indeed, Heron formula shows that

$$area(a,b,c) = \frac{1}{4} \sqrt{[(a+b+c)] \cdot [(a+b-c)(a-b+c)(-a+b+c)]}$$

What's interesting is that the radicand is the product of two symmetric polynomials. The fact that it's a product (and therefore not as "pure") motivates me to look at the following expression:

$$f(a,b,c,d) = (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)$$

Notice that $f$ is a symmetric polynomial, with $f(a,b,c,0)$ recovering the "impure" product.

Question: Does $f$ calculate anything? In other words, is there any known combinatorial or geometrical meaning of $f$? I will be glad to see if it deforms the area of a triangle in some interesting way, but any interpretation is welcome.

Solution 1:

Another interpretation is that it is the radicand in brahmagupta's formula for the area for a cyclic quadrilateral.