(x+y+z)的平方-(x-y-z)的平方
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=2x(2y+2z)
=4x(y+z)
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解：原式=[(x+y+z)+(x-y-z)][(x+y=z)-(x-y-z)]
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=2x*(2y+2z)
=4x(y+z)
利用平方差公式，得
=[(x+y+z)+(x-y-z)][(x+y+z)-(x-y-z)]
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=2x(2y+2z)
=4x(y+z)
原式=(x y z x-y-z)(x y z-x y z)
=2x(2y 2z)
=4x(y z)
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x+y+z)^2-(x-y-z)^2 =[（x+y+z)+(x-y-z) ][（x+y+z)-(x-y-z) ] =(x+y+z+x-y-z)(x+y+z-x+y+z) =(2x)*(2y+2z) =4x(y+z)
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