1. **State the problem:** Simplify the expression $$\left(\frac{1}{2} a^2 b - 1\right)\left(\frac{1}{2} a^2 b + 1\right) + \frac{3}{4} a^4 b^2 - (a^4 - 1)(b^2 + 1)$$. 2. **Use the difference of squares formula:** Recall that $$(x - y)(x + y) = x^2 - y^2$$. 3. **Apply the formula to the first product:** Let $x = \frac{1}{2} a^2 b$ and $y = 1$, so $$\left(\frac{1}{2} a^2 b - 1\right)\left(\frac{1}{2} a^2 b + 1\right) = \left(\frac{1}{2} a^2 b\right)^2 - 1^2 = \frac{1}{4} a^4 b^2 - 1.$$ 4. **Rewrite the expression:** $$\frac{1}{4} a^4 b^2 - 1 + \frac{3}{4} a^4 b^2 - (a^4 - 1)(b^2 + 1).$$ 5. **Combine like terms:** $$\frac{1}{4} a^4 b^2 + \frac{3}{4} a^4 b^2 = \cancel{\frac{1}{4}} a^4 b^2 + \cancel{\frac{3}{4}} a^4 b^2 = \frac{4}{4} a^4 b^2 = a^4 b^2.$$ 6. **Expand the last product:** $$(a^4 - 1)(b^2 + 1) = a^4 b^2 + a^4 - b^2 - 1.$$ 7. **Substitute back:** $$a^4 b^2 - 1 + a^4 b^2 - (a^4 b^2 + a^4 - b^2 - 1).$$ 8. **Distribute the minus sign:** $$a^4 b^2 - 1 + \frac{3}{4} a^4 b^2 - a^4 b^2 - a^4 + b^2 + 1.$$ 9. **Simplify terms:** - Combine $a^4 b^2$ terms: $a^4 b^2 - a^4 b^2 = 0$. - Combine constants: $-1 + 1 = 0$. So the expression reduces to: $$- a^4 + b^2.$$ **Final answer:** $$\boxed{-a^4 + b^2}.$$