$\text{a) }{x}^3=0$  

$x=0$

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

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

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$\text{b) }{x}^3-9x=0$  

$x\left(x^2-9\right)=0$

$x\left(x+3\right)\left(x-3\right)=0$  

$x=0\text{lub}x=-3\text{lub}x=3$  

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

$\frac{4x^2-12x}{x^3-9x}=\frac{4x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\frac{4}{x+3}$

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$\text{c) }{x}^3-1=0$  

$x^3=1$

$x=1$  

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

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

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$\text{d) }{x}^2+8x+7=0$

$\Delta =8^2-4\cdot 1\cdot 7=64-28=36,\sqrt{\Delta }=6$    

$x=\frac{-8-6}{2}=-7\text{lub}x=\frac{-8+6}{2}=-1$  

$\left(x+7\right)\left(x+1\right)=0$

$x=-7\text{lub}x=-1$  

$D=\mathbb{R}\backslash \left\{-7,-1\right\}$

$\frac{x^2+2x+1}{x^2+8x+7}=\frac{{\left(x+1\right)}^2}{\left(x+7\right)\left(x+1\right)}=\frac{x+1}{x+7}$

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$\text{e) }{x}^2+2x-3=0$

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

$x=\frac{-2-4}{2}=-3\text{lub}x=\frac{-2+4}{2}=1$  

$\left(x+3\right)\left(x-1\right)=0$

$x=-3\text{lub}x=1$  

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

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

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$\text{f) }{x}^2-7x+6=0$

$\Delta ={\left(-7\right)}^2-4\cdot 1\cdot 6=49-24=25,\sqrt{\Delta }=5$  

$x=\frac{7-5}{2}=1\text{lub}x=\frac{7+5}{2}=6$  

$\left(x-1\right)\left(x-6\right)=0$

$x=1\text{lub}x=6$  

$D=\mathbb{R}\backslash \left\{1,6\right\}$

$\frac{x^3-x^2-x+1}{x^2-7x+6}=\frac{x^2\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)\left(x-6\right)}=\frac{\left(x-1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x-6\right)}=\frac{x^2-1}{x-6}$

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$\text{g) }{x}^3+3x^2-4=0$

$x^3-x^2+4x^2-4=0$  

$x^2\left(x-1\right)+4\left(x^2-1\right)=0$  

$x^2\left(x-1\right)+4\left(x+1\right)\left(x-1\right)=0$  

$\left(x-1\right)\left(x^2+4\left(x+1\right)\right)=0$  

$\left(x-1\right)\left(x^2+4x+4\right)=0$  

$\left(x-1\right){\left(x+2\right)}^2=0$  

$x=1\text{lub}x=-2$  

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

$\frac{x^2+4x+4}{x^3+3x^2-4}=\frac{{\left(x+2\right)}^2}{\left(x-1\right){\left(x+2\right)}^2}=\frac{1}{x-1}$

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$\text{h) }{x}^2-49=0$

$\left(x-7\right)\left(x+7\right)=0$  

$x=-7\text{lub}x=7$  

$D=\mathbb{R}\backslash \left\{-7,7\right\}$

$\frac{x^2-7x}{x^2-49}=\frac{x\left(x-7\right)}{\left(x-7\right)\left(x+7\right)}=\frac{x}{x+7}$

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$\text{i) }{x}^2-3x=0$

$x\left(x-3\right)=0$  

$x=0\text{lub}x=3$  

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

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