jmc

Let $P$ be the polynomial defined by $P(x)=x^6-x^5+x^4-x^3+x^2-x+1$. Note that $(x+1)P(x)=x^7+1$. So the roots of $P$ are on the unit circle. Hence the roots of each quadratic factor $x^2+b_{k}x+c_k$ are also on the unit circle. Because each quadratic factor has real coefficients, its roots come in conjugate pairs. Because the roots are on the unit circle, each $c_k$ is $1$. When we expand the product of the three quadratic factors, we get a polynomial of the form $$x^6+(b_1+b_2+b_3)x^5+\cdots$$Because the coefficient of $x^5$ in $P$ is $-1$, we see that $b_1+b_2+b_3=-1$. So we have $$b_1{c}_1+b_2{c}_2+b_3{c}_3=b_1+b_2+b_3=\boxed{-1}$$.