Udowodnij, że liczba [latex] sqrt[3]{9+ sqrt{80} } + sqrt[3]{9- sqrt{80} } [/latex] jest równa 3

[latex]x=sqrt[3]{9+sqrt{80} } + sqrt[3]{9- sqrt{80} }=3[/latex]   [latex]x=sqrt[3]{9+ sqrt{80} } + sqrt[3]{9- sqrt{80} }[/latex] [latex]a=sqrt[3]{9+ sqrt{80} }[/latex] [latex]b= sqrt[3]{9-sqrt{80} }[/latex] [latex]x=a+b[/latex] -------------------------- [latex]a^3+b^3=left(sqrt[3]{9+sqrt{80} } ight) ^3+left(sqrt[3]{9- sqrt{80} } ight) ^3=[/latex] [latex]9+sqrt{80}+9- sqrt{80}=18[/latex] [latex]ab=sqrt[3]{9+sqrt{80} }cdot sqrt[3]{9- sqrt{80}} =[/latex] [latex]ab=sqrt[3]{(9+sqrt{80})(9-sqrt{80})}=81-80 =1[/latex] -------------------------- [latex](a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)[/latex] [latex](a+b)^3=18+3cdot 1(a+b)[/latex] [latex](a+b)^3=18+3(a+b)[/latex] Podstawiamy: [latex]x=a+b[/latex] [latex]x^3=18+3x[/latex] [latex]x^3-3x-18=0[/latex] [latex]x^3-27-3x+9=0[/latex] [latex](x - 3)(x^2 + 3x + 9)-3(x-3)=0[/latex] [latex](x-3)(x^2+3x+9-3)=0[/latex] [latex](x-3)(x^2+3x+6)=0[/latex] [latex]x-3=0 Rightarrow x=3[/latex] lub [latex]x^2+3x+6=0[/latex] [latex]Delta=3^2-4cdot 1cdot 6=9-24=-15<0[/latex]

I sposób [latex]sqrt[3]{9+ sqrt{80}} + sqrt[3]{9- sqrt{80}} =sqrt[3]{9+ 4sqrt{5}} + sqrt[3]{9- 4sqrt{5}} =\\ =sqrt[3]{(frac{3+sqrt{5}}{2})^3} +sqrt[3]{(frac{3-sqrt{5}}{2})^3} =frac{3+sqrt{5}}{2}+frac{3-sqrt{5}}{2}= \\ =frac{3+sqrt{5}+3-sqrt{5}}{2}= frac{6}{2} =3[/latex] cbdu ------------------- Wyjaśnienia: [latex](frac{3+sqrt{5}}{2})^3 = frac{(3+sqrt{5})^3}{2^3}=frac{3^3+3 cdot 3^2 cdot sqrt{5}+3 cdot 3 cdot (sqrt{5})^2+ (sqrt{5})^3}{8}= \\ =frac{27+27sqrt{5}+45+ 5sqrt{5}}{8}=frac{72+ 32sqrt{5}}{8}=9+ 4sqrt{5} = 9+ sqrt{80} \\\ (frac{3-sqrt{5}}{2})^3 = frac{(3-sqrt{5})^3}{2^3}=frac{3^3-3 cdot 3^2 cdot sqrt{5}+3 cdot 3 cdot (sqrt{5})^2- (sqrt{5})^3}{8}= \\ =frac{27-27sqrt{5}+45- 5sqrt{5}}{8}=frac{72-32sqrt{5}}{8}=9-4sqrt{5} = 9- sqrt{80} [/latex] ------------------- II sposób [latex]x = sqrt[3]{9+ sqrt{80}} + sqrt[3]{9- sqrt{80}} /^3 \\ x^3 = (sqrt[3]{9+ sqrt{80}} + sqrt[3]{9- sqrt{80}})^3 \\ x^3 = (sqrt[3]{9+ sqrt{80}})^3 + (sqrt[3]{9- sqrt{80}})^3+3 cdot (sqrt[3]{9+ sqrt{80}}) cdot (sqrt[3]{9- sqrt{80}}) \ cdot (sqrt[3]{9+ sqrt{80}} + sqrt[3]{9+ sqrt{80}}) \\ x^3 = 9+ sqrt{80} +9-sqrt{80}+3 cdot sqrt[3]{(9+ sqrt{80})(9- sqrt{80})} cdot x \\ x^3 = 18 +3xsqrt[3]{81-80}[/latex] [latex]x^3 = 18 +3xsqrt[3]{1} \\ x^3 = 18 +3x cdot 1 \\ x^3 = 18 +3x \\ x^3-3x - 18 = 0 \\ (x - 3)(x^2+3x+6) = 0 \\ x - 3 = 0 vee x^2 + 3x + 6 = 0 \\ x - 3 = 0 \ x = 3 \\ x^2 + 3x + 6 = 0 \ Delta = 3^2 - 4 cdot 1 cdot 6 = 9 - 24 = - 15 extless 0 \ r'ownanie nie ma rozwiaza'n \\ Zatem: \ x = 3 \\ sqrt[3]{9+ sqrt{80}} + sqrt[3]{9- sqrt{80}} = 3[/latex] cbdu ------------------- Wyjaśnienia: [latex](a + b)^3 = a^3 + 3a^2b+3ab^2 +b^3 = a^3 + b^3 + 3ab(a + b) \\\ x^3 - 3x -18 =_{dla x = 3}= 3^3 - 3 cdot 3 - 18 = 27 - 9 - 18 = 0 \ Zatem: \ (x^3 - 3x -18) : (x - 3) = x^2 + 3x +6 \ underline{-x^3 +3x^2} \ 3x^2 -3x - 18 \ underline{-3x^2+9x} \ 6x - 18 \ underline{-6x+18} \ 0 \\ x^3 - 3x -18=(x - 3)(x^2 + 3x +6)[/latex] -------------------