X=(Y-A)(Z+B) is it possible to figure X,Y, and Z if only A and B are known Given the following formula.
1,000,000 = 1,000 x 1,000

If I was to subtract A from one of the values I get B in return for example.
1,000,000 = (1,000 - 800)(1,000 + 4,000)

or even more results
1,000,000 = (1,000 - 750)(1,000 + 3,000) = (1,000 - 500)(1,000 + 1,000)

If this is expressed as a formula X=(Y-A)(Z+B) and I knew the A and B values is it possible to figure our the XYZ values? such as
X = (Y - 750)(Z + 3,000) = (Y - 500)(Z + 1,000)

 A: No, not with only one or two different pairs $A,B$.
Your final equation is equivalent to $X=YZ-750Z+3000Y-2250000=YZ-500Z+1000Y-500000$. Neglecting $X$ for the moment, we have $2000Y=250Z+1750000$. There are lots of (infinitely many) pairs $Y,Z$ which satisfy this, for example $Z=40,Y=880$ or $Z=120, Y=890$.
For each of these, the two expressions $(Y-750)(Z+3000)$ and $(Y-500)(Z+1000)$ are equal to each other, and you can just take their common value to be $X$. Here we have
$$395200=(880-750)(40+3000)=(880-500)(40+1000)\\
436800=(890-750)(120+3000)=(890-500)(120+1000)$$
If you had three expressions instead of just two, these would yield simultaneous equations for $Y$ and $Z$ which will usually only have one solution.
A: This is how you started

Given the following formula.
1,000,000 = 1,000 x 1,000
If I was to subtract A from one of the values I get B in return for example.
1,000,000 = (1,000 - 800)(1,000 + 4,000)

What you really did was provide values for $A, X, Y,$ and $Z$ and then solve for $B$. Yes, you can do that. But, if you want $B$ to be an integer, only certain values of $A, X, Y,$ and $Z$ will work.
Mathematically, it would look like this
\begin{align}
   X &= (Y-A)(Z+B) \\
   \dfrac{X}{(Y-A)} &= Z-B \\
   \dfrac{X}{(Y-A)}-Z &= -B \\
   Z - \dfrac{X}{(Y-A)} &= B
\end{align}
The rest of your question confuses me.

or even more results
X = (Y-A)(Z + B) = (Y - A')(Z + B')
If this is expressed as a formula X=(Y-A)(Z+B) and I knew the A and B values
is it possible to figure our the XYZ values? such as
X = (Y - 750)(Z + 3,000) = (Y - 500)(Z + 1,000)
If this is expressed as a formula X=(Y-A)(Z+B) and I knew the A and B values

By this I think you meant the expressions

1,000,000 = (1,000 - 750)(1,000 + 3,000) = (1,000 - 500)(1,000 + 1,000)

where you clearly provided values for $X, Y,$ and $Z$. and found two solutions $(A, B) = (750, 3000)$ and $(A, B) = (500, 1000)$. There could be other solutions, but I didn't try to find them.
I can't give an exact answer, but, yes, some of the time you can assign values to some of the variables and then find one or more solutions for the remaining variables. The number of solutions you find depends on your domain - the set of numbers that you will draw your solutions from. Like integers, postive integers, fractions, or real numbers.
Added 9/21/2021
$$ X = (Y-A)(Z + B)$$
If you are given $A$ and $B$, and you have picked a particular value of $X$...
Find a pair of integers, say $U$ and $V$ for which $UV = X$. Then $Y=U+A$ and $Z = V-B$.
