a=√3 b=-2√15 c=√3 Δ=b²-4ac=(-2√15)²-4·√3·√3=60-12=48>0 są pierwiastki [latex]x_{1}+x_{2}= -frac{b}{a} \ \ x_{1}x_{2}= frac{c}{a} [/latex] [latex]x_{1}^{2}+x_{2}^{2}=x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}-2x_{1}x_{2}=(x_{1}+x_{2})^2-2x_{1}x_2} \ \ x_{1}^{3}+x_{2}^{3}=(x_{1}+x_{2})(x_{1}^{2}-x_{1}x_{2}+x_{2}^{2}) = \ \ =(x_{1}+x_{2})((x_{1}+x_{2})^{2}-3x_{1}x_{2})[/latex] [latex]= frac{-b}{a}(( frac{-b}{a})^{2}-3 frac{c}{a})= \ \ = frac{2 sqrt{15} }{ sqrt{3} }(( frac{ 2sqrt{15}}{ sqrt{3}})^{2}-3 frac{ sqrt{3}}{ sqrt{3}})=2 sqrt{5}(20-3)=34 sqrt{5} [/latex]
[latex]f(x)=sqrt{3}x^{2}-2sqrt{15}+sqrt{3}\ \ x_{1}^{3}+x_{2}^{3}=\ \ =(x_{1}+x_{2})(x_{1}^{2}-x_{1}x_{2}+x_{2}^{2})=\ \ =(x_{1}+x_{2})(x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}-3x_{1}x_{2})=[/latex] [latex]=(x_{1}+x_{2})[(x_{1}+x_{2})^{2}-3x_{1}x_{2}]=\ \ =frac{-b}{a}*[(frac{-b}{a})^{2}-3*frac{c}{a}]=\ \ =frac{2sqrt{15}}{sqrt{3}}*[(frac{2sqrt{15}}{sqrt{3}})^{2}-3*frac{sqrt{3}}{sqrt{3}}]=\ \ =frac{2sqrt{5}*sqrt{3}}{sqrt{3}}*(frac{4*15}{3}-3*1)=\ \ =2sqrt{5}*(frac{60}{3}-3)=\ \ =2sqrt{5}*(20-3)=\ \ =2sqrt{5}*17=\ \ =34sqrt{5}[/latex]
