Simplify the Exponential Expression: (a^2x/a^y) × (a^2y/a^-y)

The problem involves multiplication between two fractions, so first we'll apply the rule for multiplying fractions which states that multiplication between two fractions is calculated by putting one fraction over a line by multiplying the numerators together and multiplying the denominators together, mathematically:

$\frac{x}{y}\cdot\frac{w}{m}=\frac{x\cdot w}{y\cdot m}$

Let's apply it to the problem:

$\frac{a^{2x}}{a^{y}}\cdot\frac{a^{2y}}{a^{-y}}=\frac{a^{2x}\cdot a^{2y}}{a^y\cdot a^{-y}}$

Next, we'll notice that both in the numerator separately and in the denominator separately there is multiplication between terms with identical bases, so we'll use the power law for multiplying terms with identical bases:

$a^m\cdot a^n=a^{m+n}$Let's apply it separately to the numerator and denominator in the problem:

$\frac{a^{2x}\cdot a^{2y}}{a^y\cdot a^{-y}}=\frac{a^{2x+2y}}{a^{y+(-y)}}=\frac{a^{2x+2y}}{a^0}$

Next, we'll remember that any number to the power of 0 equals 1, mathematically:

$a^0=1$

So let's return to the problem:

$\frac{a^{2x+2y}}{a^{0}}=\frac{a^{2x+2y}}{1}=a^{2x+2y}$

Where we actually used the fact that division by 1 doesn't change the value of the number, meaning mathematically:

$X:1=\frac{X}{1}=X$

Now let's examine the result we got above:

$a^{2x+2y}\text{ }$

In terms of simplification using the laws of exponents we have indeed finished since this is the most simplified expression,

but it's worth noting that in the exponent we got an expression that can be factored using common factor extraction:

$2x+2y=2(x+y)$

In this case the common factor is the number 2,

Let's return to the result of the expression simplification, we got:

$a^{2x+2y}=a^{2(x+y)}$Therefore the correct answer is C.