If $w,w^2$ are non-real cube roots of unity and $a,b,c \in R$ such that $$\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}=2 w^2$$ and
$$\frac{1}{a+w^2}+\frac{1}{b+w^2}+\frac{1}{c+w^2}=2 w$$ then find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$$
Could some provide some method to solve this question. I tried adding two equation to get $-2$ on R.H.S. but each term on L.H.S. is not heading towards $a+1$ type in denominator. How to generate $\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}$ ?
We can say that $$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x} = \frac{2}{x}$$ has $x=\omega\;,\omega^2$ its roots.
Now Simplifying
$$x\bigg((b+x)(c+x)+(c+x)(a+x)+(a+x)(b+x)\bigg) = 2(a+x)(b+x)(c+x)$$
$$\bigg(3x^3+2(a+b+c)x^2+(ab+bc+ca)x\bigg) = 2\bigg(x^3+(a+b+c)x^2+(ab+bc+ca)x+abc\bigg)$$
So $$x^3+0 \cdot x^2-(ab+bc+ca)x-abc = 0$$
So it is a cubic equation whose two roots are $x=\omega\;, \omega^2$
Let third root is $x\;,$ Then $x+\omega+\omega^2 = 0\Rightarrow x = 1$
So $x=1$ is third root of $$\frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x} = \frac{2}{x}$$