Question

一道初二因式分解计算9（a+1)^2(a-1)^2-6(a^2-1)(b^2-1)+(b+1)^2(b-1)^2

Answer 1

9（a+1)^2(a-1)^2-6(a^2-1)(b^2-1)+(b+1)^2(b-1)^2

=[3(a^2-1)]^2-6(a^2-1)(b^2-1)+(b^2-1)^2

=(3a^2-3)^2-6(a^2-1)(b^2-1)+(b^2-1)^2

=[(3a^2-3)-(b^2-1)]^2

=(3a^2-b^2-2)^2