|
|
Let a,b, and c be the roots of the cubic
x3+px2+qx+r = 0
Then p =-(a+b+c) = 0.
Since  a = 0  a3 = 3abc = 3 (given)
Hence abc =1.
This gives r =-1
Hence the cubic is x3+qx-1 = 0 satisfied by a,b and c.
Hence a3+qa-1 = 0 etc.
Hence a5+qa3-a2 = 0
Hence a5+qa3-a2 = 0
Now a2 = -2ab = -2q
Hence 0+3q+2q = 0
Hence q =0
The eqn is x3-1 = 0, which has the roots 1,  , 2
Hence a2007+b2007+c2007 = 1++2 = 0
PS: Nice work conjurer. I remembered that but I was half-way thru anyways. Learning slows with age
|
Moderation Team
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
836
