If $\omega$ represents a cube root of unity and $\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{k}+\frac{1}{\mathrm{\omega }}\right)\left(\mathrm{k}+\frac{1}{{\mathrm{\omega }}^2}\right)=340, \mathrm{then} \mathrm{n}=$

$\begin{array}{rcl}\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{k}+\frac{1}{\mathrm{\omega }}\right)\left(\mathrm{k}+\frac{1}{{\mathrm{\omega }}^2}\right)& =& 340\end{array}$

$\begin{array}{rcl}\sum _{\mathrm{k}=1}^{\mathrm{n}} \left(\mathrm{k}+{\mathrm{w}}^2\right) \left(\mathrm{k}+\mathrm{w}\right)& =& 340\end{array}$

$\begin{array}{rcl}\sum _{\mathrm{k}=1}^{\mathrm{n}} {\mathrm{k}}^2+\mathrm{k}(\mathrm{w}+{\mathrm{w}}^2)+{\mathrm{w}}^3& =& 340\end{array}$

$\begin{array}{rcl}\sum _{\mathrm{k}=1}^{\mathrm{n}} \left({\mathrm{k}}^2-\mathrm{k}+1\right)& =& 340\end{array}$

$\begin{array}{rcl} & =& \frac{\mathrm{n}(\mathrm{n}+1) (2\mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}=340\left(6\right)\end{array}$

$\begin{array}{rcl} & & \\ \mathrm{n}\left(2{\mathrm{n}}^2+3\mathrm{n}+1-3\mathrm{n}-3+6\right)& =& 2040\\ \mathrm{n}\left(2{\mathrm{n}}^2+4\right)& =& 2040\\ 2\mathrm{n}\left({\mathrm{n}}^2+2\right)& =& 2\times \left(1020\right)\\ {\mathrm{n}}^3+2\mathrm{n}& =& {10}^3+2\left(10\right) \Rightarrow \boxed{ \mathrm{n}=10}\end{array}$