If $\mathrm{\alpha }$ is a root of ${\mathrm{Z}}^2-\mathrm{Z}+1=0$ then

$$\begin{gathered} \left({\mathrm{\alpha }}^{2014}+\frac{1}{{\mathrm{\alpha }}^{2014}}\right)+{\left({\mathrm{\alpha }}^{2015}+\frac{1}{{\mathrm{\alpha }}^{2015}}\right)}^2+ \\ {\left({\mathrm{\alpha }}^{2016}+\frac{1}{{\mathrm{\alpha }}^{2016}}\right)}^3+{\left({\mathrm{\alpha }}^{2017}+\frac{1}{{\mathrm{\alpha }}^{2017}}\right)}^4+ \\ {\left({\mathrm{\alpha }}^{2018}+\frac{1}{{\mathrm{\alpha }}^{2018}}\right)}^5= \end{gathered}$$

$\begin{array}{rcl}{\mathrm{z}}^2-\mathrm{z}+1& =& 0 \Rightarrow \mathrm{z}=-\mathrm{w} \mathrm{or} - {\mathrm{w}}^2\end{array}$

$\begin{array}{rcl}\mathrm{Let} \mathrm{\alpha }& =& -\mathrm{w}\end{array}$

$\begin{array}{rcl}\mathrm{\alpha }& =& -\mathrm{w} \Rightarrow {\mathrm{\alpha }}^{2014}={\mathrm{w}}^{2014}={\left({\mathrm{w}}^3\right)}^{671}·\mathrm{w}=\mathrm{w}\end{array}$

$\begin{array}{rcl} {\mathrm{\alpha }}^{2015}& =& -{\mathrm{w}}^2, {\mathrm{\alpha }}^{2016}={\mathrm{w}}^3=1, {\mathrm{\alpha }}^{2017}=-\mathrm{w}, {\mathrm{\alpha }}^{2018}={\mathrm{w}}^2\end{array}$

$\begin{array}{rcl} & & \left({\mathrm{\alpha }}^{2014}+\frac{1}{{\mathrm{\alpha }}^{2014}}\right)+{\left({\mathrm{\alpha }}^{2015}+\frac{1}{{\mathrm{\alpha }}^{2015}}\right)}^2+{\left({\mathrm{\alpha }}^{2016}+\frac{1}{{\mathrm{\alpha }}^{2016}}\right)}^3+{\left({\mathrm{\alpha }}^{2017}+\frac{1}{{\mathrm{\alpha }}^{2017}}\right)}^4+{\left({\mathrm{\alpha }}^{2018}+\frac{1}{{\mathrm{\alpha }}^{2018}}\right)}^5\end{array}$

$\begin{array}{rcl} & & \therefore \left(\mathrm{w}+\frac{1}{\mathrm{w}}\right)+{\left(-{\mathrm{w}}^2-\frac{1}{{\mathrm{w}}^2}\right)}^2+{\left(1+\frac{1}{1}\right)}^3+{\left(-\mathrm{w}-\frac{1}{\mathrm{w}}\right)}^4+{\left({\mathrm{w}}^2+\frac{1}{{\mathrm{w}}^2}\right)}^5\\ & =& -1+1^2+2^3+1-1=8\end{array}$