Solution: Given $ \fracab = 3 $, let $ a = 3b $. Substituting into $ a + b = 16 $: $ 3b + b = 16 $, so $ b = 4 $ and $ a = 12 $. Now, $ a^2 - b^2 $ factors as $ (a - b)(a + b) = (12 - 4)(16) = 8 \times 16 = 128 $. Therefore, the value is $ \boxed128 $. - Abu Waleed Tea