jmc

Rewriting the original equation: $$\begin{array}{rl}{x}^3-10x^2+25x& >0\\ \Rightarrow x(x^2-10x+25)& >0\\ \Rightarrow x(x-5{)}^2& >0\end{array}$$If $x<0,$ then $x(x-5{)}^2<0,$ and if $x=0,$ then $x(x-5{)}^2=0.$

If $0<x<5,$ then $x(x-5{)}^2>0.$ If $x=5,$ then $x(x-5{)}^2=0.$ If $x>5,$ then $x(x-5{)}^2>0.$ Therefore, the solution is $$x\in \boxed{(0,5)\cup (5,\infty )}.$$