Is there a way to simplify this equation?

Question

$$ A = \left( 4000 \left( 1+\frac{x}{y} \right) \right)^4 \cdot \left( 1 + \frac{x+0.002}{y} \right)^4 \cdot \left(1+\frac{x+0.002+0.002}{y} \right)^4 $$

Answer 1

$$ A = \left( 4000 \left( 1+\frac{x}{y} \right) \right)^4 \cdot \left( 1 + \frac{x+\dfrac{2}{1000}}{y} \right)^4 \cdot \left(1+\frac{x+\dfrac{4}{1000}}{y} \right)^4 $$ Then, $$ A = \left( 4000 \left( 1+\frac{x}{y} \right) \cdot \left( 1 + \frac{x+\dfrac{2}{1000}}{y} \right) \cdot \left(1+\frac{x+\dfrac{4}{1000}}{y} \right)\right)^4 $$ Then, $$ A = \left(\dfrac{256}{y^{12}}\right)\cdot10^{-12}\cdot\left( \left( y+x \right) \cdot \left( 1000\cdot y+1000\cdot x+2 \right) \cdot \left(1000\cdot y+1000\cdot x+4 \right)\right)^4 $$ Then, $$ A = \left(\dfrac{256}{y^{12}}\right)\cdot10^{-12}\cdot\left( 10^6(x+y)^3+6\cdot 10^3(x+y)^2+8(x+y)\right)^4 $$ Then, $$ A = \left(\dfrac{256}{y^{12}}\right)\cdot10^{-12}\cdot(x+y)^4\cdot\left( 10^6(x+y)^2+6\cdot 10^3(x+y)+8\right)^4 $$

Thus, putting $x+y=X$ and $X_1=4\cdot 10^{-3}$ and $X_2=2\cdot 10^{-3}$:

$$ A = \left(\dfrac{256}{y^{12}}\right)\cdot10^{-12}\cdot(X)^4\left( 10^6X^2+6\cdot 10^3X+8\right)^4 $$ Therefore, $$ A = \left(\dfrac{256}{y^{12}}\right)\cdot10^{12}\cdot X^4\left( (X+X_1)\cdot(X+X_2)\right)^4 $$ Thus, $$ A = 256\cdot\left(\dfrac{10}{y}\right)^{12}\cdot (x+y)^4\cdot (x+y+X_1)^4\cdot(x+y+X_2)^4 $$ Or, $$ A = 256\cdot\left(\dfrac{1}{10y}\right)^{12}\cdot (x+y)^4\cdot (1000x+1000y+2)^4\cdot(1000x+1000y+4)^4 $$