Question

If $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }+\frac{1}{d+\omega }=\frac{2}{\omega }$ where $a$, $b$, $c$ are real and $\omega$ is a non-real cube root of unity, then

• A. $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}+\frac{1}{d+{\omega }^2}=\frac{2}{{\omega }^2}$
• B. $abc+bcd+abd+acd=4$
• C. $a+b+c+d=-2abcd$
• D. $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$

Answer 1

Answer: (D)

$\frac{1}{a+\omega }+\frac{1}{b+\omega }\frac{1}{c+\omega }\frac{1}{d+\omega }=\frac{2}{\omega }$
$\Rightarrow \omega$ is the root of the equation
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}$
$\therefore 2x^4+\left(\Sigma a\right)x^2+0.x^2-\left(\Sigma abc\right)x-2abcd=0$
Let $\alpha ,\beta ,\gamma$ be the other roots, then
$\omega +\alpha +\beta +\gamma =-\frac{\Sigma a}{2}\dots \left(1\right)$
$\Sigma \alpha \beta =0$
i.e., $\alpha \beta +\alpha \omega +\beta \omega +\gamma \omega +\beta \gamma +\gamma \alpha =0\dots \left(2\right)$
$\Sigma \alpha \beta \gamma =\frac{\Sigma abc}{2}\dots \left(3\right)$
$\alpha \beta \gamma \omega =-abcd\dots \left(4\right)$
$\therefore \alpha ,\beta +\omega .\left(\alpha +\beta +\gamma \right)+\beta \gamma +\gamma \alpha =0$
[By $2$]
Since complex roots are always in conjugate form
$\gamma =\bar{\omega }={\omega }^2\left(\bar{\omega }={\omega }^2\therefore 1+\omega +{\omega }^2=0\right)$
$\alpha \beta +\omega \left(\alpha +\beta \right)+\omega .{\omega }^2+{\omega }^2\left(\alpha +\beta \right)0$
$\Rightarrow \alpha \beta +\left(\omega +{\omega }^2\right)\left(\alpha +\beta \right)+{\omega }^3=0$
$\Rightarrow \alpha \beta +\left(-1\right)\left(\alpha +\beta \right)+1=0$
$\Rightarrow \alpha \beta -\alpha -\beta +1=0$
$\Rightarrow \alpha \left(\beta -1\right)-\left(\beta -1\right)=0$
$\Rightarrow \left(\alpha -1\right)\left(\beta -1\right)=0$
$\therefore$ either $\alpha =1$ or $\beta =1$
Hence one root is unity.
$\therefore \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$