x²-y²-z²=0,x³-y³-z³=(x-y)(x-z)
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x²-y²-z²=0,x³-y³-z³=(x-y)(x-z)A.问:A是关于xyz的一次多项
这个是(x-y)乘以(x-z)乘以A 明天12:00之前有效!
这个是(x-y)乘以(x-z)乘以A 明天12:00之前有效!
∵ ∴
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
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
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