Question

If $\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }+\frac{1}{d+\omega }=2{\omega }^2$
and $\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}+\frac{1}{d+{\omega }^2}=2\omega$
then prove that
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}=2$

Answer 1

Because ${\omega }^2=\frac{1}{\omega }$ and hence the two questions can be written as
$\sum \frac{1}{a+\omega }=\frac{2}{\omega },\sum \frac{1}{a+{\omega }^2}=\frac{2}{{\omega }^2}$ and
we have prove that $\sum \frac{1}{a+1}=\frac{2}{1}$
Consider the equation $\sum \frac{1}{a+x}=\frac{2}{x}$ .......(1)
whose two roots are $\omega ,{\omega }^2$ by virtue of given equations and we have to show x = 1 is also a root of this equation
$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}$
$\therefore \sum x(x+a)(x+b)(x+c)=2(x+a)(x+b)(x+c)(x+d)$
or $\sum x[x^1+x^2(a+b+c)+x(ab+bc+ca)+abc]$
$=z[x^4+x^3\underset{4}{\omega }a+x^2\omega \underset{6}{a}b+x\omega \underset{4}{a}bc+\underset{1}{abcd}]$
$4x^4+x^3\underset{12}{\left(3\omega a\right)}+x^{2}2\underset{12}{\omega ab}+x\omega \underset{4}{abc}$
$=2[x^4+x^3\omega a+x^2\omega ab+x\omega abc+abcd]$
$\therefore 2x^4+x^3\omega a+o{x}^2-x\omega abc-2bacd=0$
Above is a fourth degree equation two of whose roots are $\omega$ and ${\omega }^2$ and let the other two be $\alpha ,\beta$ then
$\alpha +\beta +\omega +{\omega }^2=-\frac{\sum a}{z}$
or $\alpha +\beta -1=-\frac{\sum a}{2}$ ...........(2)
$\because \omega +{\omega }^2+1=0,{\omega }^3=1$
Products taken two at a time = 0 as coefficient of $x^2$ is zero
$(\alpha +\beta )(\omega +{\omega }^2)+\alpha \beta ++\omega .{\omega }^2=0$
or $(\alpha +\beta )-(-1)+\alpha \beta +1=0$
$\alpha \beta -(\alpha +\beta )+1=0$
$\therefore (\alpha -1)(\beta -1)=0$
$\therefore$ if $\alpha$ = 1 then $\beta =-\frac{\sum a}{2}$ and if $\beta$ = 1 then
$\alpha =-\frac{\sum a}{2}$ by..(2)
Hence in either case the other two roots are 1 and $-\frac{\sum a}{2}$.Putting x = 1 in (1), we get the required result