Question

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)

Answer 1

你好

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)
=[x+（y+z）] [(y+z）-x] [x-（y-z)] [x+（y-z)]
=[(y+z）²-x²][x²-（y-z）²]
=(y+z）²x²-(y+z）²（y-z）²-x⁴+x²（y-z）²
=x²[(y+z）²+（y-z）²]-（y²-z²）²-x⁴
=x²（2y²+2z²）-y⁴+2y²z²-z⁴-x⁴
=-x⁴-y⁴-z⁴+2x²y²+2x²z²+2y²z²

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Answer 2

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)
=[(y+z)+x][(y+z)-x][x-(y-z)][x+(y-z)]
=[(y+z)²-x²][x²-(y-z)²]
=[(y²+2yz+z²)-x²][x²-(y²-2yz+z²)]
=(y²+2yz+z²-x²)(x²-y²+2yz-z²)
=y²(x²-y²+2yz-z²)+2yz(x²-y²+2yz-z²)+z²(x²-y²+2yz-z²)-x²(x²-y²+2yz-z²)
=x²y²-y^4+2y³z-y²z²+2x²yz-2y³z+4y²z²-2yz³+x²z²-y²z²+2yz³-z^4-x^4+x²y²-2x²yz+x²z²
=2x²y²-y^4+2y²z²+2x²z²-z^4-x^4

Answer 3

把前两个分别一组，后两个分成一组
（x+y+z）（-x+y+z）＝（y+z）＾2-X＾2
（x-y+z）（x+y-z）＝-（z-y）＾2+x＾2

Answer 4

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)
=[x+（y+z）] [(y+z）-x] [x-（y-z)] [x+（y-z)]
=[(y+z）²-x²][x²-（y-z）²]
=(y+z）²x²-(y+z）²（y-z）²-x⁴+x²（y-z）²
=x²[(y+z）²+（y-z）²]-（y²-z²）²-x⁴
=x²（2y²+2z²）-y⁴+2y²z²-z⁴-x⁴
=-x⁴-y⁴-z⁴+2x²y²+2x²z²+2y²z²

Answer 5