$\text{a)}{x}^3-4x^2+4x=0$  

$x\left(x^2-4x+4\right)=0$  

$x{\left(x-2\right)}^2=0$  

$x=0\vee x-2=0$  

$x=0\vee x=2$  

Zatem: 

$D=\mathbb{R}\backslash \left\{0,2\right\}$   

 

$\text{b)}9x^3-6x^2+x=0$  

$x\left(9x^2-6x+1\right)=0$  

$x{\left(3x-1\right)}^2=0$  

$x=0\vee 3x-1=0$  

$x=0\vee 3x=1\mid :3$  

$x=0\vee x=\frac{1}{3}$  

Zatem: 

$D=\mathbb{R}\backslash \left\{0,\frac{1}{3}\right\}$  

 

$\text{c)}3x^3+x^2-2x=0$  

$x\left(3x^2+x-2\right)=0$  

$x=0\vee \underset{\left(\star \right)}{\underbrace{3x^2+x-2=0}}$  

Wyznaczamy pierwiastki równania  $\left(\star \right)$ . 

$3x^2+x-2=0$  

$\Delta =1^2-4\cdot 3\cdot \left(-2\right)=1+24=25$  

$\sqrt{\Delta }=\sqrt{25}=5$  

$x_1=\frac{-1-5}{2\cdot 3}=\frac{-6}{6}=-1$  

$x_2=\frac{-1+5}{2\cdot 3}=\frac{4}{6}=\frac{2}{3}$  

Zatem: 

$D=\mathbb{R}\backslash \left\{-1,0,\frac{2}{3}\right\}$  

 

$\text{d)}{x}^3+3x^2+3x+1=0$  

${\left(x+1\right)}^3=0$  

$x+1=0$

$x=-1$   

Zatem: 

$D=\mathbb{R}\backslash \left\{-1\right\}$  

 

$\text{e)}{x}^3-6x^2+12x-8=0$  

${\left(x-2\right)}^3=0$  

$x-2=0$

$x=2$    

Zatem: 

$D=\mathbb{R}\backslash \left\{2\right\}$  

 

$\text{f)}{x}^3-64=0$  

$x^3=64$

$x=\sqrt[3]{64}$

$x=4$    

Zatem: 

$D=\mathbb{R}\backslash \left\{4\right\}$