When will a parametric solution exist for a Diophantine equation? Many Diophantine equations with infinite solutions I've seen have parametric solution. Example:
$$a=m^2-n^2$$
$$b=2mn$$
$$c=m^2+n^2$$
Implies: $$a^2+b^2=c^2$$
So pythagorean triples can be generated with arbitrary $$m>n>0$$
I specifically have in mind this equation:
$$ab(a+b)(a-b)=cd(c+d)(c-d)$$
So does there exist some f such that:
$$f(r, s, t)=(a,b,c,d)$$
 A: Equation, ab(a+b)(a-b)=cd(c+d)(c-d) has parametrization and is given below,
a=(k)(5k-1)
b=(3k-1)(7k-2)
c=(2k-1)(3k-1)
d=(5k-1)(4k-1)
for k=3 we have,
(a,b,c,d)=(21,76,20,77)
A: Your equation can be expressed in the form,
$$ab(a^2-b^2)=cd(c^2-d^2)\tag1$$
More generally,
$$ab(a^2+hb^2)=cd(c^2+hd^2)\tag2$$
This is the expanded version of the well-known,

$$(a + b \sqrt{h})^4 + (c - d \sqrt{h})^4 = (a - b \sqrt{h})^4 + (c + d \sqrt{h})^4\tag3$$

Yours was the case $\color{blue}{h=-1}$ so, in effect, you are looking for solutions  in the complex numbers to $(3)$. We borrow the method used by Euler. Let,
$$a,\;b,\;c,\;d = x,\; p y,\; q x,\; y$$
to get,
$$(p - q^3) x^2 = -h (p^3 - q) y^2$$
implying
$$(p - q^3)(p^3 - q) =-h\,z^2\tag4$$
This is an elliptic curve and if you find one rational point, you can find infinitely many. 
I. For $h=1$, so $(3)$ is in the integers:
$$q = p + \frac{3p (1 - p^2)^3}{4p^6 + p^4 + 10p^2 + 1}$$
II. For $h=-1$, so $(3)$ is in the Gaussian integers:
$$q = -p - \frac{3p (1 + p^2)^3}{-4p^6 + p^4 - 10p^2 + 1}$$
From these initial formulas, one can then find infinitely more.

Example. Let $h=-1$ and $p=2$, then,
$$a,\;b,\;c,\;d = 186,\; 178,\; 128,\; 89$$
which solves your $(1)$ and implying,
$$(186 + 178\,i)^4 + (128 - 89 \,i)^4 = (186 - 178 \,i)^4 + (128 + 89 \,i)^4$$
