高一数学 分数指数幂化简题 题如图

先分开来看：
第一个分式=[a^(1/3)](a-8b)/[4b^(2/3)+2(ab)^(1/3)+a^(2/3)]
第二个分式=1/a^(2/3)-[2b^(1/3)]/a=[a-2a(2/3)b^(1/3)]/a^(5/3)=[a^(1/3)-2b^(1/3)]/a
第三个分式=[a^(5/6)]/[a^(1/6)]=a^(2/3)
故原式={[a^(1/3)](a-8b)/[4b^(2/3)+2(ab)^(1/3)+a^(2/3)]}×[a/[a^(1/3)-2b^(1/3)]×a^(2/3)
=[a²(a-8b)]/{[4b^(2/3)+2(ab)^(1/3)+a^(2/3)][a^(1/3)-2b^(1/3)]}
=[a²(a-8b)]/(a-8b)=a²