由x-y=6,xy=-8
(x+y)^2=(x-y)^2+4xy=36-32=4,得到x+y=2或者-2,现将x+y=2代入=½(x+y+z)²+½(x-y-z)(x-y+z)-z(x+y) ==½(2+z)²+½(6-z)(6+z)-z(2) =20
将x+y=-2代入=½(x+y+z)²+½(x-y-z)(x-y+z)-z(x+y) ==½(-2+z)²+½(6-z)(6+z)-z(-2) =20
故不论怎样,该代数式=20