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
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}$
$\therefore 2x^{4}+\left(\Sigma a\right)x^{2}+0. x^{2}-\left(\Sigma\,abc\right)x-2 \,abcd=0$
Let $\alpha, \beta, \gamma$ be the other roots, then
$\omega+\alpha+\beta+\gamma=-\frac{\Sigma a}{2}\quad\ldots\left(1\right)$
$\Sigma\alpha\beta=0$
i.e., $\alpha\beta+\alpha\omega+\beta\omega+\gamma\omega+\beta\gamma+\gamma\alpha=0\,\ldots\left(2\right)$
$\Sigma\alpha\beta\gamma=\frac{\Sigma\,abc}{2}\quad\ldots\left(3\right)$
$\alpha\,\beta\,\gamma\omega=-abcd\quad\ldots\left(4\right)$
$\therefore \alpha, \beta+\omega. \left(\alpha+\beta+\gamma\right)+\beta\gamma+\gamma\alpha=0$
[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.
$\therefore \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$