"Roots" would imply that
(x^3 + 125) / (x +5) = 0
Factoring the top as a sum of cubes, we get
(x + 5)(x^2 - 5x + 25) So we have
(x + 5)(x^2 - 5x + 25) / (x + 5) = 0 Only the numerator can make this zero
So setting both factors to 0, we have
(x + 5 ) = 0 so x =-5, but we must reject this since it makes the denominator of the original function undefined.
The other two roots are non-real...using the on-site solver, we have
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{5}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{5}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.330\: \!127\: \!018\: \!925}}{i}\right)\\
{\mathtt{x}} = {\frac{{\mathtt{5}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4.330\: \!127\: \!018\: \!925}}{i}\\
\end{array} \right\}$$
So...we have no real solutions to this "rational" function
