<s>[INST] Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$. 
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
  The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.

At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6} 
 & = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
 & = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
 & = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\ 
 & = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
 & = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\ 
\end{align}
And 
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.
 [/INST] We have that $q$ is a root of
$$
X^6 +X^5+\dots +X+1=X^3(X^3+X^{-3} +X^{2}+X^{-2}+ X+X^{-1}+1).
$$
Hence $q+q^{-1}$ is a root of
$$
Y^3 - 3Y +Y^2 -2 +Y +1= Y^3+Y^2-2Y-1.
$$
Hence $\frac{1}{q+q^{-1}}$ is a root of 
$$
Z^3+2Z-Z-1.
$$
Exactly the same is true for the roots $q^2, q^4$, so we get that the sum of the three roots 
$$\frac{1}{q+q^{-1}}+\frac{1}{q^2+q^{-2}}+\frac{1}{q^4+q^{-4}}=-2.$$
The left hand side is equal to the expression we are to evaluate. 
</s>