If $a_i\in\mathbb{R}$, $\omega^2+\omega+1=0$, and $\sum_{i=1}^n\frac{1}{a_i+\omega^k} =2\omega^{2k}$ for $k=1,2$, find $\sum_{i=1}^n\frac{1}{a_i+1}$. 
In this question, $\omega$ is the complex cube root of $1$ and $a_i \in \mathbb R$.   If $$\sum_{i=1}^n \frac{1}{a_i + \omega} =2\omega ^2$$ and $$\sum_{i=1}^n \frac{1}{a_i + \omega ^2} =2\omega\,,$$
then find $$\sum_{i=1}^n \frac{1}{a_i + 1}\,.$$

I think I have solved it correctly (answer is $2$) , but please check my proof.
Proof:
Let $f: \mathbb C \to \mathbb C$ such that $$f(x)=\left (\sum_{i=1}^n \frac{1}{a_i + x}\right) -2x^2 $$
As per the given condition, $\omega$ and $\omega ^2$ are the roots of the equation $f(x)=0$.
Now, if we have some equation whose roots are $\alpha , \beta$ (for example) then by using the transformation $x \to x^{1/3}$, we get the equation whose roots are $\alpha ^3, \beta ^3$.
Using that here, $f(x^{1/3})=0$ should have the roots $\omega ^3, \omega ^6 $ that is,
$1$ is a root of $f(x^{1/3})=0$
$\implies f(1)=0$
$$\implies \sum_{i=1}^n \frac{1}{a_i + 1}=2$$
Which is the right answer, by the way.
My question:
Is this proof using transformations correct or not?
 A: The problem, as pointed out, in the comments is that the "transformation" $x \mapsto x^{1/3}$ is multivalued and you're misusing that.
Let us see this better.
When you wish to make the substitution $x \mapsto x^{\color{red}{1/3}}$, you should make a choice of which cube root you are choosing for each $x$.
Suppose you have made the choice to choose $\omega$ for $x = 1$. This means that you must fix $1^{\color{red}{1/3}} = \omega$ in your later substitutions.
This is the error that you made. Using the fact that $\omega$ is a root, you concluded that $1$ is a root of $f(x^{\color{red}{1/3}})$ but then you must remember that $1^{\color{red}{1/3}}$ is $\omega$.
In particular, you cannot write
$$f(1^{\color{red}{1/3}}) = f(1) = 0$$
since $1^{\color{red}{1/3}} = \omega$ and not $1$.

Similarly, if you considered $x \mapsto x^{\color{blue}{1/3}}$ and chose $\omega^2$ for $1^{\color{blue}{1/3}}$, you'd run into the same problem.

I already gave an example in the comments of how you could use this to derive that $1$ is a root of $x^2 + x + 1$ even though it's not.
Another example of such a thing is misuse of the square root.
For example, you could consider $g(x) = x+1$, note that $-1$ is a root and then using the transformation $x \mapsto \sqrt{x}$ incorrectly, conclude that $1$ is also a root.
(In fact, this is the spirit behind many "proofs" of $-1=1$.)
