Question

因式分解：（x²-y²）²-8（x²+y²）+16. 谢谢!？

Answer 1

（x²-y²）²-8（x²+y²）+16
=(x²-y²)² -8(x²-y²)+16-16y²
=(x²-y²-4)²-16y²
=(x²-y²-4+4y)(x²-y²-4-4y)
=[x²-(y²-4y+4)][x²-(y²+4y+4)]
=[x²-(y-2)²][x²-(y+2)²]
=(x+y-2)(x-y+2)(x+y+2)(x-y-2),6,（x²-y²）²-8（x²+y²）+16
=(x²-y²)² -8(x²-y²)+16-16y²
=(x²-y²-4)²-16y²
=(x²-y²-4+4y)(x²-y
...,1,原式=(x^2+y^2)^2-8(x^2+y^2)+16-4x^2*y^2
=(x^2+y^2-4)^2-(2xy)^2
=(x^2+y^2+2xy-4)(x^2+y^2-2xy-4)
=[(x+y)^2-2^2][(x-y)^2-2^2]
=(x+y-2)(x+y+2)(x-y+2)(x-y-2)
不懂，请追问，祝愉快O(∩_∩)O~,0,