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 $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$

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

Answer 1

The correct option is A $2$

Given relation show that $\omega$ ${\omega }^2$ 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 $3x^2+2x^2\sum a+x\sum bc$ $=2x^3+2x^2\sum a+2x\sum ab+2abc$
$x^3+0x^2-x\sum ab-2abc=0$
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 $\therefore \sum \frac{1}{a+1}=\frac{2}{1}$