题目内容：

a=4的三次方根+2的三次方根+1的三次方根,求3/a + 3/a*2 + 3/a*3

优质解答

a=4的三次方根+2的三次方根+1的三次方根
=(2^2)^(1/3)+2^(1/3)+1
=2^(2/3)+2^(1/3)+1 ①
∴3/a + 3/a*2 + 3/a*3=3/a + 3/a*2 + 1/a
=4/a+3/2a=11/2a
=(11/2)×(1/a) ②
把①代入②,得
3/a + 3/a*2 + 3/a*3
=(11/2)×{1/[2^(2/3)+2^(1/3)+1]}
=(11/2)×[2^(1/3)-1]/{[2^(2/3)+2^(1/3)+1][2^(1/3)-1]}
=(11/2)×[2^(1/3)-1]/(2-1)
=11[2^(1/3)-1]/2