送分題－拆立方根

已知(√27+√28)^1/3+(√27－√28)^1/3=√3
求(√27+√28)^1/3-(√27－√28)^1/3
Sol
設 A=(√27+√28)^1/3，B=(√27－√28)^1/3
A+B=√3
AB=[(√27+√28)(√27－√28)]^1/3=－1
A^3－B^3=4√7
再設 A－B=P>0
P^3=A^3－3A^2B+3AB^2－B^3
=(A^3－B^3)－3AB(A－B)
=4√7+3P
P^3－3P－4√7=0
(P^2－P√7+4)(P－√7)=0
P^2－P√7+4，D=7－4*1*4<0
So P^2－P√7+4>0
So
P－√7=0
P=√7

001,你抄漏咗個4

[(√27 + √28)^(1/3) - (√27 - √28)^(1/3)]²

= 3 - "4"[(√27 + √28)(√27 - √28)]^(1/3)

(√27 + √28)^(1/3) + (√27 - √28)^(1/3) = √3 [(√27 + √28)^(1/3) + (√27 - √28)^(1/3)]² = (√3)² (√27 + √28)^(2/3) + 2[(√27 + √28)^(1/3)][(√27 - √28)^(1/3)]+ (√27 - √28)^(2/3) = 3 (√27 + √28)^(2/3) - 2[(√27 + √28)^(1/3)][(√27 - √28)^(1/3)]+ (√27 - √28)^(2/3) = 3 - 4[(√27 + √28)^(1/3)][(√27 - √28)^(1/3)] [(√27 + √28)^(1/3) - (√27 - √28)^(1/3)]²= 3 - [(√27 + √28)(√27 - √28)]^(1/3)= 3 - (27 -28)^(1/3)= 3 - (-1)^(1/3)= 3 - (-1) (complex root rej.)= 4 (√27 + √28)^(1/3) - (√27 - √28)^(1/3) = 2 I HOPE THIS CAN HELP YOU! ^_^

2012-06-24 20:50:46 補充：
Too Careless :(