因式分解(x³+y³)²-9x²y²(x+y)²
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因式分解(x³+y³)²-9x²y²(x+y)²
原式
=(x³+y³)²-[3xy(x+y)]²
=[(x³+y³)+3xy(x+y)][(x³+y³)-3xy(x+y)]
=[(x+y)(x²-xy+y²)+3xy(x+y)][(x+y)(x²-xy+y²)-3xy(x+y)]
=(x+y)(x²-xy+y²+3xy)(x+y)(x²-xy+y²-3xy)
=(x+y)²(x²+2xy+y²)(x²-4xy+y²)
=(x+y)^4(x²-4xy+y²)
再问: 您的回答很好,但是从(x³+y³)²-[3xy(x+y)]²是如何变成[(x³+y³)+3xy(x+y)][(x³+y³)-3xy(x+y)]的呢
再答: 这是平方差公式 a^2-b^2=(a+b)(a-b)
=(x³+y³)²-[3xy(x+y)]²
=[(x³+y³)+3xy(x+y)][(x³+y³)-3xy(x+y)]
=[(x+y)(x²-xy+y²)+3xy(x+y)][(x+y)(x²-xy+y²)-3xy(x+y)]
=(x+y)(x²-xy+y²+3xy)(x+y)(x²-xy+y²-3xy)
=(x+y)²(x²+2xy+y²)(x²-4xy+y²)
=(x+y)^4(x²-4xy+y²)
再问: 您的回答很好,但是从(x³+y³)²-[3xy(x+y)]²是如何变成[(x³+y³)+3xy(x+y)][(x³+y³)-3xy(x+y)]的呢
再答: 这是平方差公式 a^2-b^2=(a+b)(a-b)
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