Question

Hello...

Answer 1

Given\[\frac{1}{a+\omega} + \frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega} = 2 \omega^2\]\[\frac{1}{a+\omega^2} + \frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2} = 2 \omega\]Prove\[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2\]

Answer 2

If I replace \(\omega=1\) I get the thesis

Answer 3

Here, \(\omega\) is a cube root of unity.

Answer 4

I don't knew, sorry!

Answer 5

we can rewrite the second equation as below:
\[\Large \frac{1}{{a + \bar \omega }} + \frac{1}{{b + \bar \omega }} + \frac{1}{{c + \bar \omega }} + \frac{1}{{d + \bar \omega }} = \frac{2}{{\bar \omega }}\]
where:
\[\Large {\omega ^2} = \bar \omega ,\quad {\omega ^3} = 1\]

Answer 6

indeed... then?

Answer 7

one root is \(\omega=1\), so, we have:
\[\Large \omega = 1 \Rightarrow \bar \omega = 1\]

Answer 8

@ganeshie8