Question

几道因式分解 （1）x³+x²+2xy+y²+y³（2）x³+3x&s

Answer 1

(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)