求详细过程!!!
1个回答
展开全部
原式=(a^(1/3)-b^(1/3))(a^(2/3)+a^(1/3)b^(1/3)+b^(2/3))/(a^(1/3)-b^(1/3))
-(a^(1/3)+b^(1/3))(a^(2/3)-a^(1/3)b^(1/3)+b^(2/3))/(a^(1/3)+b^(1/3))
=(a^(2/3)+a^(1/3)b^(1/3)+b^(2/3))-(a^(2/3)-a^(1/3)b^(1/3)+b^(2/3))
=a^(1/3)b^(1/3)+a^(1/3)b^(1/3)
=2a^(1/3)b^(1/3)
-(a^(1/3)+b^(1/3))(a^(2/3)-a^(1/3)b^(1/3)+b^(2/3))/(a^(1/3)+b^(1/3))
=(a^(2/3)+a^(1/3)b^(1/3)+b^(2/3))-(a^(2/3)-a^(1/3)b^(1/3)+b^(2/3))
=a^(1/3)b^(1/3)+a^(1/3)b^(1/3)
=2a^(1/3)b^(1/3)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询