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几道因式分解
(1)x³+x²+2xy+y²+y³
(2)x³+3x²y+3xy²+y³
(3)x³-3x²y+3xy²-y³
(4)x³+3x²+3x+2
人气:329 ℃ 时间:2020-04-10 18:51:52
解答
(1)(x³+y³)+ (x²+2xy+y²)=(x+y)(x²-xy+y²)+(x+y)(x+y)=(x+y)(x²-xy+y²+x+y)
(2)(x³+y³)+3xy(x+y)=(x+y)(x²-xy+y²)+3xy(x+y)=(x+y)(x+y)(x+y)
(3)(x³-y³)-3xy(x-y)=(x-y)(x²+xy+y²)-3xy(x-y)=(x-y)(x-y)(x-y)
(4)(x³-1)+3x²+3x+3=(x-1)(x²+x+1)+3(x²+x+1)=(x+2)(x²+x+1)
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