W zadaniu korzystamy ze wzorów skróconego mnożenia. 

• ${\left(a+b\right)}^2=a^2+2ab+b^2$  
  
  
• ${\left(a-b\right)}^2=a^2-2ab+b^2$  
  
  
• $\left(a-b\right)\left(a+b\right)=a^2-b^2$  

 

Mamy więc: 

$\text{a)}\left(2-3\sqrt{3}\right)\left(2-3\sqrt{3}\right)={\left(2-3\sqrt{3}\right)}^2=2^2-2\cdot 2\cdot 3\sqrt{3}+{\left(3\sqrt{3}\right)}^2=4-12\sqrt{3}+27=31-12\sqrt{3}$  

$\text{b)}\left(7+5\sqrt{2}\right)\left(5\sqrt{2}+7\right)=\left(7+5\sqrt{2}\right)\left(7+5\sqrt{2}\right)={\left(7+5\sqrt{2}\right)}^2=7^2+2\cdot 7\cdot 5\sqrt{2}+{\left(5\sqrt{2}\right)}^2=$  

$=49+70\sqrt{2}+50=99+70\sqrt{2}$  

 

$\text{c)}\left(4\sqrt{5}+\sqrt{3}\right)\left(4\sqrt{5}-\sqrt{3}\right)={\left(4\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2=80-3=77$  

 

$\text{d)}\left(\sqrt{8}+2\sqrt{6}\right)\left(2\sqrt{6}-\sqrt{8}\right)=\left(2\sqrt{6}+\sqrt{8}\right)\left(2\sqrt{6}-\sqrt{8}\right)={\left(2\sqrt{6}\right)}^2-{\left(\sqrt{8}\right)}^2=24-8=16$     

 

$\text{e)}{\left(\sqrt{6}-3\sqrt{3}\right)}^2={\left(\sqrt{6}\right)}^2-2\cdot \sqrt{6}\cdot 3\sqrt{3}+{\left(3\sqrt{3}\right)}^2=6-6\sqrt{6\cdot 3}+27=33-6\sqrt{18}=$    

$=33-6\sqrt{9\cdot 2}=33-6\cdot 3\sqrt{2}=33-18\sqrt{2}$  

 

$\text{f)}{\left(2\sqrt{32}-3\sqrt{2}\right)}^2={\left(2\sqrt{32}\right)}^2-2\cdot 2\sqrt{32}\cdot 3\sqrt{2}+{\left(3\sqrt{2}\right)}^2=128-12\sqrt{32\cdot 2}+18=146-12\sqrt{64}=$   

$=146-12\cdot 8=146-96=50$