解
x³+3x²y+3xy²+y³
=(x³+y³)+(3x²y+3xy²)
=(x+y)(x²-xy+y²)+3xy(x+y)
=(x+y)(x²-xy+y²+3xy)
=(x+y)(x²+y²+2xy)
=(x+y)³
熟悉的话
可以直接得到
=(x+y)³
而(x-y)³=x³-3x²y+3xy²-y³