If  $\alpha ,\beta$ are roots of the equation $x^2-4x-3=0,$  then $\frac{{\alpha }^{-\alpha \beta }({\alpha }^{\alpha }{\alpha }^{\beta }-4{\alpha }^{-\alpha \beta })+{\beta }^{-\alpha \beta }({\beta }^{\alpha }{\beta }^{\beta }-4{\beta }^{-\alpha \beta })}{\alpha .{\alpha }^{\alpha +\beta }+\beta .{\beta }^{\alpha +\beta }}=$

$$\begin{gathered} \frac{{\alpha }^{-\alpha \beta }({\alpha }^{\alpha }{\alpha }^{\beta }-4{\alpha }^{-\alpha \beta })+{\beta }^{-\alpha \beta }({\beta }^{\alpha }{\beta }^{\beta }-4{\beta }^{-\alpha \beta })}{\alpha .{\alpha }^{\alpha +\beta }+\beta .{\beta }^{\alpha +\beta }} \\ \frac{{\alpha }^{\alpha +\beta -\alpha \beta }+{\beta }^{\alpha +\beta -\alpha \beta }-4({\alpha }^{-2\alpha \beta }+{\beta }^{-2\alpha \beta })}{{\alpha }^{\alpha +\beta +1}+{\beta }^{\alpha +\beta +1}} \\ \alpha +\beta =4,\alpha \beta =-3 \\ \frac{{\alpha }^7+{\beta }^7-4({\alpha }^6+{\beta }^6)}{{\alpha }^5+{{\rm B}}^5} \\ {S}_n={\alpha }^n+{\beta }^n \\ {S}_n-4S_{n-1}-3S_{n-2}=0 \\ {S}_7-4S_6-3S_5=0 \end{gathered}$$