Question

求答案！急急急急急急急！

Answer 1

解：原式＝[﹙x＋y﹚²－z²][z²－﹙x－y﹚²]
＝﹙2xy＋z²＋x²＋y²﹚﹙2xy＋z²－x²－y²﹚
＝﹙2xy＋z²﹚²－﹙x²＋y²﹚²
＝4x²y²＋4xyz²＋z^4－x^4－2x²y²－y^4.
＝2x²y²＋4xyz²＋z^4－x^4－y^4.

追问

x^4这代表啥？

追答

x^4表示x的四次方。

Answer 2

原式=(（y+z)+x）((y+z)-x)(x-(y-z))(x+(y-z))
=((y+z)^2-x^2)(x^2-(y-z)^2)
=(y+z)^2*x^2-x^4-(y+z)^2*(y-z)^2+x^2*(y-z)^2
=x^2*((y+z)^2+(y-z)^2)-x^4-((y+z)(y-z))^2
=x^2*(y^2+2yz+z^2+y^2-2yz+z^2)-x^4-((y^2-z^2))^2
=x^2*(2y^2+2z^2)-x^4-(y^4-2y^2*z^2+z^4)
=2x^2*y^2+2x^2*z^2-x^4-y^4+2y^2*z^2-z^4
=-x^4-y^4-z^4+2x^2*y^2+2x^2*z^2+2y^2*z^2

Answer 3