$b)$  

$x^2-3x-18\ne 0$  

$\Delta ={\left(-3\right)}^2-4\cdot 1\cdot \left(-18\right)=9+72=81$  

$\sqrt{\Delta }=9$  

$x_1=\frac{3-9}{2}=\frac{-6}{2}=-3$  

$x_2=\frac{3+9}{2}=\frac{12}{2}=6$  

$\underline{\underline{x\in \mathbb{R}\backslash \left\{-3,6\right\}}}$  

 

 

 

$\frac{x^2+6x+9}{x^2-3x-18}=\frac{{\left(x+3\right)}^2}{\left(x+3\right)\left(x-6\right)}=\frac{x+3}{x-6}$   

 

 

 

 

$c)$  

$x^2+2x-3\ne 0$  

$\Delta =2^2-4\cdot 1\cdot \left(-3\right)=4+12=16$  

$\sqrt{\Delta }=4$  

$x_1=\frac{-2-4}{2}=\frac{-6}{2}=-3$  

$x_2=\frac{-2+4}{2}=\frac{2}{2}=1$  

$\underline{\underline{x\in \mathbb{R}\backslash \left\{-3,1\right\}}}$  

 

$\frac{x^3-x}{x^2+2x-3}=\frac{x\left(x^2-1\right)}{\left(x+3\right)\left(x-1\right)}=\frac{x\sout{\left(x-1\right)}\left(x+1\right)}{\left(x+3\right)\sout{\left(x-1\right)}}=\frac{x\left(x+1\right)}{x+3}$