Question

Розкласти на множники x³+y³+3x²-3xy+3y²

Answer 1

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)