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

Answer 1

The correct option is B $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$

Since, $\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac{2}{w}$

Thus $w$ is the root of the equation.

$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}$

$\Rightarrow 2x^4+(\sum a)x^3+0.x^2-(\sum abc)x-2abcd=0$

Let $\alpha ,\beta ,\gamma$ be the roots of this equation, so

$w+\alpha +\beta +\gamma =-\frac{\sum a}{2}$............(1)

$\sum ab=0$

$\alpha \beta +\alpha w+\beta w+\gamma w+\gamma \beta +\alpha \gamma =0$...................(2)

$\sum \alpha \beta \gamma =\frac{\sum abc}{2}$......(3)

$\alpha \beta \gamma w=-abcd$........4

Now since complex roots occur in conjugate pairs,

$\gamma =\overline{w}=w^2$

Therefore from (2), we have,

$\alpha \beta +\alpha w+\beta w+w^{2}w+w^2\beta +\alpha w^2=0$

$\alpha \beta +(\alpha +\beta )(w^2+w)+w^3=0\Rightarrow \alpha \beta +(\alpha +\beta )(-1)+1=0\Rightarrow (\alpha -1)(\beta -1)=0$

So $\alpha =1$ or $\beta =1$ one root is unity,

Thus, $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$