jmc

First we factor $x$ from the numerator, $$\frac{x(1-10x+25x^2)}{8-x^3}.$$Now we see the square of a binomial in the numerator, so our expression is equal to $$\frac{x(1-5x{)}^2}{8-x^3}.$$The denominator only has the single (real) root $x=2$, and we can make more sense of that by applying the difference of cubes factorization $$\frac{x(1-5x{)}^2}{(2-x)(x^2+2x+4)}.$$Now we can factor the entire rational function as $$\left(\frac{x}{2-x}\right)\left(\frac{(1-5x{)}^2}{x^2+2x+4}\right).$$Note that the denominator $x^2+2x+4=(x+1{)}^2+3$ is always positive. The factor $x$ changes sign at $x=0,$ the factor $2-x$ changes sign at $x=2,$ and the factor $1-5x$ changes sign at $x=\frac{1}{5}.$ We build a sign chart accordingly.

$$\begin{array}{ccccc}& x<0& 0<x<\frac{1}{5}& \frac{1}{5}<x<2& 2<x\\ x& -& +& +& +\\ 2-x& +& +& +& -\\ (1-5x{)}^2& +& +& +& +\\ \left(\frac{x}{2-x}\right)\left(\frac{(1-5x{)}^2}{x^2+2x+4}\right)& -& +& +& -\end{array}$$Also, the expression $$\left(\frac{x}{2-x}\right)\left(\frac{(1-5x{)}^2}{x^2+2x+4}\right)$$is equal to 0 at $x=0$ and $x=\frac{1}{5},$ so the solution to $$\left(\frac{x}{2-x}\right)\left(\frac{(1-5x{)}^2}{x^2+2x+4}\right)\ge 0$$is $x\in \boxed{[0,2)}.$