Question

If $\omega$ be complex cube root of unity satisfying the equation
$\frac{1}{a+\omega }+\frac{1}{b+\omega }+\frac{1}{c+\omega }=2{\omega }^2$ and
$\frac{1}{a+{\omega }^2}+\frac{1}{b+{\omega }^2}+\frac{1}{c+{\omega }^2}+=2\omega$ , then
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ is equal to

• A. $2$
• B. $-2$
• C. - 1 + ${\omega }^2$
• D. - 1 + $\omega$

Answer 1

The correct option is A $2$

Given relations show that $\omega$ and ${\omega }^2$ are the roots of $\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}=\frac{2}{x}$ which is a cubic in x and we have to prove that 1 is also a root of it. Above equation is simplified to
$x\sum (b+x)(c+x)=2(a+x)(b+x)(c+x)$
or $x[\sum bc+2)[\sum a]x+3x^2]=2[x^3+x^2\sum a+x\sum ab+abc]$
or $3x^3+2x^2\sum a+x\sum bc=2x^3+2x^2\sum a+2x\sum ab+2abc$
$x^3+0x^2-x\sum ab-2abc=0$
Above is a cubic in x whose two roots are $\omega$, ${\omega }^2$. If $\alpha$ be the third root then $\alpha +\omega +{\omega }^2$ = sum of the roots = 0 as the term of $x^2$ is missing. Put $\omega +{\omega }^2$ = - 1
$\therefore$ $\alpha$ - 1 = 0 or $\alpha$ = 1 is the third root
$\therefore \sum \frac{1}{a+1}=\frac{2}{1}$