Factorization of (a^2-b^2)(c^2-d^2)-4abcd

Dear Learners, I present an easy method to solve the given problem.
(a^2-b^2)(c^2-d^2)-4abcd
=a^2(c^2-d^2)-b^2(c^2-d^2)-4abcd, using distributive property in this and the next step
=a^2c^2-a^2d^2-b^2c^2+b^2d^2-4abcd
=a^2c^2+b^2d^2-2abcd-a^2d^2-b^2c^2-2abcd, using commutativity
=(a^2c^2+b^2d^2-2abcd)-(a^2d^2+b^2c^2+2abcd), using (A+B)^2=A^2+2AB+B^2,etc.
=(ac-bd)^2-(ad+bc)^2
=(ac-bd+ad+bc)(ac-bd-ad-bc) Ans.
To arrive at the last step, I used the property A^2-B^2=(A+B)(A-B)

Factorization of x^6+63x^3-64

Solution: Let x^3=a
So, x^6+63x^3-64
=(x^3)^2+63x^3-64,using A^MN=(A^M)^N
=a^2+63a-64,by substitution
=a^2+64a-a-64
=a(a+64)-1(a+64)
=(a+64)(a-1),taking (a+64) as common
=(x^3+64)(x^3-1),substituting back
=(x^3+4^3)(x^3-1^3)
=(x+4)(x^2-4x+4^2)(x-1)(x^2+1x+1^2),using A^3+B^3=(A+B)(A^2-AB+B^2),etc
=(x+4)(x^2-4x+16)(x-1)(x^2+x+1) Ans.