1. **State the problem:** Simplify the expression $$ (d^2 - c^2 + a^2 - b^2)^2 - 4(bc - da)^2 $$. 2. **Recall the formula:** This expression resembles the difference of squares formula: $$ x^2 - y^2 = (x - y)(x + y) $$ where $$ x = d^2 - c^2 + a^2 - b^2 $$ and $$ y = 2(bc - da) $$. 3. **Rewrite the expression:** $$ (d^2 - c^2 + a^2 - b^2)^2 - 4(bc - da)^2 = (x)^2 - (2(bc - da))^2 = x^2 - y^2 $$ 4. **Apply difference of squares:** $$ = (x - y)(x + y) = \big(d^2 - c^2 + a^2 - b^2 - 2(bc - da)\big) \times \big(d^2 - c^2 + a^2 - b^2 + 2(bc - da)\big) $$ 5. **Simplify each factor:** First factor: $$ d^2 - c^2 + a^2 - b^2 - 2bc + 2da $$ Group terms: $$ (d^2 + 2da + a^2) - (c^2 + 2bc + b^2) = (d + a)^2 - (c + b)^2 $$ Second factor: $$ d^2 - c^2 + a^2 - b^2 + 2bc - 2da $$ Group terms: $$ (d^2 - 2da + a^2) - (c^2 - 2bc + b^2) = (d - a)^2 - (c - b)^2 $$ 6. **Apply difference of squares again:** First factor: $$ (d + a)^2 - (c + b)^2 = \big((d + a) - (c + b)\big) \times \big((d + a) + (c + b)\big) = (d + a - c - b)(d + a + c + b) $$ Second factor: $$ (d - a)^2 - (c - b)^2 = \big((d - a) - (c - b)\big) \times \big((d - a) + (c - b)\big) = (d - a - c + b)(d - a + c - b) $$ 7. **Final factorization:** $$ (d + a - c - b)(d + a + c + b)(d - a - c + b)(d - a + c - b) $$ **Answer:** The original expression factors into the product of these four binomials.