jmc

We have that ${\omega }^3=1.$ Then ${\omega }^3-1=0,$ which factors as $(\omega -1)({\omega }^2+\omega +1)=0.$ Since $\omega$ is nonreal, ${\omega }^2+\omega +1=0.$ By the quadratic formula, $$\omega =-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i.$$Taking the conjugate of the given equation, we get $$\frac{1}{a_1+\overline{\omega }}+\frac{1}{a_2+\overline{\omega }}+\cdots +\frac{1}{a_n+\overline{\omega }}=2-5i.$$Note that if $a$ is a real number, then $$\begin{array}{rl}\frac{1}{a+\omega }+\frac{1}{a+\overline{\omega }}& =\frac{a+\omega +a+\overline{\omega }}{(a+\omega )(a+\overline{\omega })}\\ & =\frac{2a+\omega +\overline{\omega }}{a^2+(\omega +\overline{\omega })a+\omega \overline{\omega }}\\ & =\frac{2a-1}{a^2-a+1}.\end{array}$$Therefore, $$\begin{array}{rl}\sum _{k=1}^n\frac{2a_k-1}{a_k^2-a_k+1}& =\sum _{k=1}^n\left(\frac{1}{a_k+\omega }+\frac{1}{a_k+\overline{\omega }}\right)\\ & =2+5i+2-5i\\ & =\boxed{4}.\end{array}$$