若x-y=5,xy=14,则x3-x2y+xy2-y3=?
问题描述：
若x-y=5,xy=14,则x3-x2y+xy2-y3=?
答
x3-x2y+xy2-y3
=(x^2+y^2)(x-y)
=[(x-y)^2+2xy](x-y)
=[5^2+2*14]*5
=265
答
x3-x2y+xy2-y3
=x²(x-y)+y²(x-y)
=(x²+y²)(x-y)
=[(x-y)²+2xy](x-y)
=(5²+28)*5
=53*5
=265
答
解；原试=（x-y）(x^2+xy+y^2)-xy(x-y)=14*[（x-y）^2+3xy]-14*5=868
答
x-y=5，xy=14
x²＋y²＝﹙x－y﹚²＋2xy=5²＋2×14=53
x3-x2y+xy2-y3
=x²﹙x－y﹚＋y²﹙x－y﹚
=﹙x－y﹚﹙x²＋y²﹚
=5×53
=265
答
x-y=5,xy=14
x²+y²=(x-y)²+2xy=5²+2×14=53
x³-x²y+xy²-y³
=x³-y³-x²y+xy²
=(x-y)(x²+xy+y²)-xy(x-y)
=(x-y)(x²+xy+y²-xy)
=(x-y)(x²+y²)
=5×53
=265