SOLUTION: If {{{1/(a+u)+1/(b+u)+1/(c+u)+1/(d+u)=2/u}}}, {{{1/(a+v)+1/(b+v)+1/(c+v)+1/(d+v)=2/v}}}, Where u and v are imaginary cube roots of unity and a, b, c, d are constants, then prove

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: If {{{1/(a+u)+1/(b+u)+1/(c+u)+1/(d+u)=2/u}}}, {{{1/(a+v)+1/(b+v)+1/(c+v)+1/(d+v)=2/v}}}, Where u and v are imaginary cube roots of unity and a, b, c, d are constants, then prove      Log On


   

Question 1062528: If 1%2F%28a%2Bu%29%2B1%2F%28b%2Bu%29%2B1%2F%28c%2Bu%29%2B1%2F%28d%2Bu%29=2%2Fu,
1%2F%28a%2Bv%29%2B1%2F%28b%2Bv%29%2B1%2F%28c%2Bv%29%2B1%2F%28d%2Bv%29=2%2Fv,
Where u and v are imaginary cube roots of unity and a, b, c, d are constants,
then prove that
1%2F%28a%2B1%29%2B1%2F%28b%2B1%29%2B1%2F%28c%2B1%29%2B1%2F%28d%2B1%29=2,
Answer by ikleyn(52795) About Me  (Show Source):
You can put this solution on YOUR website!
.
The competent mathematician will never write
". . . imaginary cube roots of unity",

because they are not imaginary. They are COMPLEX instead.


What is the source of this problem ?


I just saw this problem once (long time ago) in this forum and worked on it.

And my conclusion was: the formulation is wrong/incorrect.

So, my question is: WHAT IS THE SOURCE OF IT ?