Suppose the roots of $P(x)=x^3+ax^2+bx+c$ are $\cos\frac{2\pi}7,\cos\frac{4\pi}7,\cos\frac{6\pi}7$. What is $a+b+c$?

1. As shown in the problem,\begin{gather*}a=-\sum\cos\tfrac{2k\pi}7=\tfrac12,\quad b=\sum(\text{pairwise products})=-\tfrac12,\\ c=-\prod\cos\tfrac{2k\pi}7=-\tfrac18.\end{gather*}
2. $a+b+c=\frac12-\frac12-\frac18=\boxed{-\dfrac18}$.