x³+y³+3x²-3xy+3y²=(x+y)(x²-xy+y)+(3(x²-xy+y))=
=(x+y)(x²-xy+y)+3(x²-xy+y)=(x²-xy+y²)(x+y+3)
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x³+y³+3x²-3xy+3y²=(x+y)(x²-xy+y)+(3(x²-xy+y))=
=(x+y)(x²-xy+y)+3(x²-xy+y)=(x²-xy+y²)(x+y+3)