∵ ∴
解∵x²-y²-z²=0
∴z²=x²-y² y²=x²-z²
x³-y³-z³=x³-y³-z(x²-y²)
=(x-y)(x²+xy+y²)-z(x-y)(x+y)
=(x-y)(x²+xy+y²-zx-zy)
又注意到 y²=x²-z²
∴ 上式=(x-y)(x²+xy+x²-z²-zx-zy)
=(x-y)(2x²-zx-z²+xy-zy)
=(x-y)[(2x²-zx-z²)+(xy-zy)]
=(x-y)[(2x-z)(x-z)+y(x-z)]
=(x-y)(x-z)(2x-z+y)
所以A=2x-z+y