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)


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.
