what are the roots of (x^3+125)/(x+5)

"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