Question

If $(a+\omega {)}^{-1}+(a+\omega {)}^{-1}+(c+\omega {)}^{-1}+(d+\omega {)}^{-1}=2{\omega }^{-1}$ and $(a+{\omega }^{\prime }{)}^{-1}+(b+{\omega }^{\prime }{)}^{-1}+(c+{\omega }^{\prime }{)}^{-1}+(d+{\omega }^{\prime }{)}^{-1}=2({\omega }^{\prime }{)}^{-1}$, where $\omega$ and ${\omega }^{\prime }$ are the imaginary cube roots of unity, then the value of $(a+1{)}^{-1}+(b+1{)}^{-1}+(c+1{)}^{-1}+(d+1{)}^{-1}$ is

Answer 1

$\because \omega$ 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.

$(a+x{)}^{-1}+(b+x{)}^{-1}+(c+x{)}^{-1}+(d+x{)}^{-1}=2x^{-1}$ ........ (1)
$\Rightarrow 2x^4+\sum a{x}^3+0\cdot x^2-\sum abcx-2abcd=0$

Let the other roots of the equation are $\alpha$ and $\beta$ then
$\alpha +\beta +\omega +{\omega }^{\prime }=-\frac{\sum a}{2}$
$\Rightarrow \alpha +\beta -1=-\frac{\sum a}{2}$
and $\sum \alpha \beta =0$ .......... (2)
$(\alpha +\beta )(\omega +{\omega }^{\prime })+\alpha \beta +\omega {\omega }^{\prime }=0$
$\Rightarrow (\alpha +\beta )(-1)+\alpha \beta +1=0$
$\Rightarrow (\alpha -1)(\beta -1)=0$
$\therefore \alpha =1$ or $\beta =1$
If $\alpha =1$, then $\beta =-\frac{\sum a}{2}$ and if $\beta =1$ then $\alpha =-\frac{\sum a}{2}$.
Hence $1$ is the root of the equation (1)
$\therefore (a+1{)}^{-1}+(b+1{)}^{-1}+(c+1{)}^{-1}+(d+1{)}^{-1}=2$
Ans: 2