(x+y+z)(x-y+z)(x+y-z)(-x+y+z)
展开全部
原式=[(x+z)+y][(x+z)-y][y+(x-z)][y-(x-z)]
=[(x+z)²-y²][y²-(x-z)²]
=(x²+2xz+z²-y²)(y²-x²+2xz-z²)
=[2xz+(x²-y²+z²)][2xz-(x²-y²+z²)]
=4x²z²-(x^4+y^4+z^4-2x²y²+2x²z²-2y²z²)
=-x^4-y^4-z^4+2x²y²+2x²z²+2y²z²
=[(x+z)²-y²][y²-(x-z)²]
=(x²+2xz+z²-y²)(y²-x²+2xz-z²)
=[2xz+(x²-y²+z²)][2xz-(x²-y²+z²)]
=4x²z²-(x^4+y^4+z^4-2x²y²+2x²z²-2y²z²)
=-x^4-y^4-z^4+2x²y²+2x²z²+2y²z²
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
