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 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. $abc+bcd+abd+acd=2$
• 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: (A), (B), (C), (D)

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

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

Taking conjugate on both the sides,
$\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}+\frac{1}{d+{\omega }^2}=\frac{2}{{\omega }^2}(\because \overline{\omega }={\omega }^2)$

So, $\omega ,{\omega }^2$ are the roots of the equation,
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}\cdots \left(1\right)$
$\Rightarrow x(4x^3+3\left(\sum a\right)x^2+2\left(\sum ab\right)x+\sum abc)$
$=2(x^4+\left(\sum a\right)x^3+\left(\sum ab\right)x^2+\left(\sum abc\right)x+abcd)$

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

Let the other $2$ roots be $\alpha ,\beta$
Here $x^2\text{coefficient}=0$
Thus $\sum \left(\text{Product of roots taken two at a time}\right)=0$

$\Rightarrow {\omega }^3+\omega \alpha +\omega \beta +{\omega }^2\alpha +{\omega }^2\beta +\alpha \beta =0$
$\Rightarrow 1-\alpha -\beta +\alpha \beta =0$
$\Rightarrow (1-\alpha )(1-\beta )=0$
$\Rightarrow \alpha =1\text{or}\beta =1$

So, $x=1$ will be the root for equation $1$.

Replacing $x=1$ in equation $\left(1\right)$, we get
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2,$
Now, if $\alpha =1$, let the other root be $\beta$

From $\left(2\right),$
Product of roots$=1\cdot \omega \cdot {\omega }^2\cdot \beta =-abcd$
$\Rightarrow \beta =-abcd[\because {\omega }^3=1]$

Sum of the roots$=-\frac{\sum a}{2}$
$\Rightarrow 1+\omega +{\omega }^2-abcd=-\frac{\sum a}{2}$
$\Rightarrow a+b+c+d=2abcd\cdots \left(3\right)$

Replacing $x=1$ in equation $\left(2\right),$ we get
$2+\sum a-\sum abc-2abcd=0$
$\Rightarrow \sum abc=2+\sum a-2abcd$
$\Rightarrow \sum abc=2\left[\text{From}\right(3\left)\right]$