If $\alpha ,\beta$ are the root of the equation $x^2-5x-1=0,$ then the value of $\frac{{\alpha }^{15}+{\alpha }^{11}+{\beta }^{15}+{\beta }^{11}}{{\alpha }^{13}+{\beta }^{13}}=$  

$x^2-5x-1=0$ 
$\alpha , \beta$ are roots then $\alpha +\beta =5,\alpha \beta =-1$   
$\frac{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{11}+{\beta }^{11}}{{\alpha }^{13}+{\beta }^{13}}=\frac{{\alpha }^{15}+{\beta }^{15}+{\alpha }^2{\beta }^2({\alpha }^{11}+{\beta }^{11})}{{\alpha }^{13}+{\beta }^{13}}$ 
$\left[\alpha \beta =-1\Rightarrow {\alpha }^2{\beta }^2=1\right]$ 
$=\frac{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{13}{\beta }^2+{\alpha }^2{\beta }^{13}}{{\alpha }^{13}+{\beta }^{13}}$ 
$=\frac{{\alpha }^{13}({\alpha }^2+{\beta }^2)+{\beta }^{13}({\beta }^2+{\alpha }^2)}{{\alpha }^{13}+{\beta }^{13}}$ 
$=\frac{({\alpha }^{13}+{\beta }^{13})({\alpha }^2+{\beta }^2)}{{\alpha }^{13}+{\beta }^{13}}$ 
$={(\alpha +\beta )}^2-2\alpha \beta$ 
$=25+2=27$