Question

If $(a+\omega {)}^{-1}+(b+\omega {)}^{-1}+(c+\omega {)}^{-1}+(d+\omega {)}^{-1}=2{\omega }^{-1},(a+\omega {)}^{-1}+(b+\omega {)}^{-1}+(c+\omega {)}^{-1}+(d+\omega {)}^{-1}=2({\omega }^{\prime }{)}^{-1}$ where $\omega$ and ${\omega }^{\prime }$ are the imaginary cube root of unity, prove that
$(a+\omega {)}^{-1}+(b+\omega {)}^{-1}+(c+\omega {)}^{-1}+(d+\omega {)}^{-1}=2$.

Answer 1

$\omega$ and ${\omega }^{\prime }$ are the imaginary cube roots of unity,
i.e., $\omega =\frac{-1+i\sqrt{3}}{2}$ and ${\omega }^{\prime }=\frac{-1-i\sqrt{3}}{2}$
$\therefore \omega +{\omega }^{\prime }=-1$ and $\omega {\omega }^{\prime }=1$
$\Rightarrow \omega$ and ${\omega }^{\prime }$ are the roots of equation.
Since, R.H.S has $\omega$ is denominator, $\omega$ must be one of component of LCM of L.H.S
This is possible only if a,b,c and d are multiples of $\omega$
Let $a=k_1\omega ,b=k_2\omega ,c=k_3\omega$ and $d=k_4\omega$
where $k_1,k_2,k_3$ and $k_4$ are arbitary constants.
(Note: a,b,c,d are also arbitary constants).
$LHS=\frac{1}{\omega (k_1+1)}+\frac{1}{\omega (k_2+1)}+\frac{1}{\omega (k_3+1)}+\frac{1}{\omega (k_4+1)}=\frac{2}{\omega }\Rightarrow \frac{1}{k_1+1}+\frac{1}{k_2+1}+\frac{1}{k_3+1}+\frac{1}{k_4+1}=2$
as $k_1,k_2,k_3$ and $k_4$ are arbitary constant, they can be replaced with a,b,c and d.
$\Rightarrow \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$.