"Plugging in" the 3 options as a strategy makes sense here.

First, you can re-write $A^8 B^4-A^4 B^2$ by factoring out $A^2$ as $(A^2)^4B^4 - (A^2)^2B^2$
Then, we substitute the options for $A^2$.

A) If $A^2 =\frac{2}{B}$ :

$(\frac{2}{B})^4B^4 - (\frac{2}{B})^2B^2$
$\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}$
$16 - 4 = 12$

Thus, A is a possible option

A) If $A^2 = -\frac{2}{B}$ :

At a quick glance, any negative value raised to an even degree (in this case, to the 2nd degree/squared) is a positive value, since the product of 2 negative values is a positive.
Thus, this will give us the same answer at Option A.

$(-\frac{2}{B})^4B^4 - (-\frac{2}{B})^2B^2$
$\dfrac{16B^4}{B^4} - \dfrac{4B^2}{B^2}$
$16 - 4 = 12$

Thus, B is a possible option

A) If $A^2 = \frac{3}{B}$ :

$(\frac{3}{B})^4B^4 - (\frac{3}{B})^2B^2$
$\dfrac{81B^4}{B^4} - \dfrac{9B^2}{B^2}$
$81 - 9 = 72$

Thus, C is NOT a possible option.