Answer
The solutions are $r=0$ or $r=-1$.
Work Step by Step
$\frac{2}{r}+\frac{1}{r^2}+\frac{r^2+r}{r^3}=\frac{1}{r}$
$\frac{2r^2+r+r^2+r}{r^3}=\frac{1}{r}$
$\frac{3r^2+2r}{r^3}=\frac{1}{r}$
$3r^3+2r^2=r^3$
$2r^3+2r^2=0$
$2r^2(r+1)$
$r=0$ or $r=-1$
Check:
$\frac{2}{0}+\frac{1}{0^2}+\frac{0^2+r}{0^3}=\frac{1}{0}$
$0=0$
$\frac{2}{-1}+\frac{1}{(-1)^2}+\frac{(-1)^2+(-1)}{(-1)^3}=\frac{1}{(-1)}$
$-1=-1$
The solutions are $r=0$ or $r=-1$.
