Matemáticas, pregunta formulada por SmithRecopilacionMat, hace 16 días

Si ∛(∛2-1)=∛a+∛b+∛c , calcular ∛(abc) ; a,b,c ∈Q

Respuestas a la pregunta

Contestado por SmithValdez
0

Analizamos \ el \  primer \ miembro\\\\\sqrt[3]{(\sqrt[3]{2}-1).\dfrac{\sqrt[3]{2}^{2}+\sqrt[3]{2}(1)+1}{\sqrt[3]{2}^{2}+\sqrt[3]{2}(1)+1}}

\sqrt[3]{\dfrac{2-1}{\sqrt[3]{2}^{2}+\sqrt[3]{2}(1)+1} }

\dfrac{1}{\sqrt[3]{\sqrt[3]{2}^{2}+\sqrt[3]{2}(1)+1}}

\dfrac{1}{\sqrt[3]{\sqrt[3]{2}^{2}+\sqrt[3]{2}(1)+1}}.\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}}

\dfrac{\sqrt[3]{3}}{\sqrt[3]{3\sqrt[3]{2}^{2}+3\sqrt[3]{2}(1)+3}}

pero \ recordemos  \ que:\\3=2+1=\sqrt[3]{2}^{3}+1

\dfrac{\sqrt[3]{3}}{\sqrt[3]{3\sqrt[3]{2}^{2}+3\sqrt[3]{2}(1)+\sqrt[3]{2}^{3}+1}}

\dfrac{\sqrt[3]{3}}{\sqrt[3]{\sqrt[3]{2}^{3}+3.\sqrt[3]{2}^{2}(1)+3\sqrt[3]{2}(1^{2})+1^{3}}}

\dfrac{\sqrt[3]{3}}{\sqrt[3]{(\sqrt[3]{2}+1)^{3}}}

\dfrac{\sqrt[3]{3}}{\sqrt[3]{2}+1}

\dfrac{\sqrt[3]{3}}{\sqrt[3]{2}+1}.\dfrac{\sqrt[3]{2}^{2}-\sqrt[3]{2}(1)+1^{2}}{\sqrt[3]{2}^{2}-\sqrt[3]{2}(1)+1^{2}}

\dfrac{\sqrt[3]{3}(\sqrt[3]{2}^{2}-\sqrt[3]{2}(1)+1^{2})}{2+1}

\dfrac{\sqrt[3]{12}-\sqrt[3]{6}+\sqrt[3]{3}}{3}

\dfrac{\sqrt[3]{12}-\sqrt[3]{6}+\sqrt[3]{3}}{\sqrt[3]{3}^{3}}

\sqrt[3]{\dfrac{12}{27}}-\sqrt[3]{\dfrac{6}{27}}+\sqrt[3]{\dfrac{3}{27}}

\sqrt[3]{\dfrac{4}{9}}+\sqrt[3]{-\dfrac{2}{9}}+\sqrt[3]{\dfrac{1}{9}}

comparando:\\\\a=\dfrac{4}{9} ;b=-\dfrac{2}{9};c=\dfrac{1}{9}

nos  \ pide  \ calcular \ :\\\\\sqrt[3]{abc}=\sqrt[3]{(\dfrac{4}{9})(-\dfrac{2}{9})(\dfrac{1}{9})}=-\dfrac{2}{9}

\mathbb{AUTOR}:  \mathrm{SmithValdez}

\mathrm{Viva \ ronaldooo!}

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