Question

Unifor-CE
Simplificando-se $(2^{- \frac{2}{3} } - 3^{-\frac{2}{3}}) . ( \sqrt[3]{3} - \sqrt[3]{2}) ^{-1} . \sqrt[3]{36}$

$a) 6^{\frac{2}{3} } \\ b)2^{\frac{1}{3} } + 3^{\frac{1}{3} }\\ c)3^{\frac{1}{3}} - 2^{\frac{1}{3} } \\ d)5^{\frac{1}{3} }\\ e)6$

gabarito: b

ajuda??

Answer 1

$(2^{-\frac{2}{3}}-3^{-\frac{2}{3}})\cdot\left(\sqrt[3]{3}-\sqrt[3]{2}\right)^{-1}\cdot\sqrt[3]{36}=\\\\\left(\dfrac{1}{\sqrt[3]{2^2}}-\dfrac{1}{\sqrt[3]{3^2}}\right)\cdot\left(\dfrac{1}{\sqrt[3]{3}-\sqrt[3]{2}}\right)\cdot\sqrt[3]{36}=\\\\\\\left(\dfrac{\sqrt[3]{3^2}-\sqrt[3]{2^2}}{\sqrt[3]{2^2}\cdot\sqrt[3]{3^2}}\right)\cdot\left(\dfrac{1}{(\sqrt[3]{3}-\sqrt[3]{2})}\right)\cdot\sqrt[3]{36}=$


$\left(\dfrac{\sqrt[3]{3^2}-\sqrt[3]{2^2}}{\sqrt[3]{6^2}}\right)\cdot\left(\dfrac{1}{(\sqrt[3]{3}-\sqrt[3]{2})}\right)\cdot\sqrt[3]{36}=\\\\\\\left(\dfrac{\sqrt[3]{3^2}-\sqrt[3]{2^2}}{(\sqrt[3]{6^2})\cdot(\sqrt[3]{3}-\sqrt[3]{2})}\right)\cdot\sqrt[3]{6^2}=\\\\\\\dfrac{(\sqrt[3]{6^2})\cdot(\sqrt[3]{3^2}-\sqrt[3]{2^2})}{(\sqrt[3]{6^2})\cdot(\sqrt[3]{3}-\sqrt[3]{2})}=\\\\\\\dfrac{(\sqrt[3]{3^2}-\sqrt[3]{2^2})}{(\sqrt[3]{3}-\sqrt[3]{2})}=$


$\dfrac{(\sqrt[3]{3}-\sqrt[3]{2})\cdot(\sqrt[3]{3}+\sqrt[3]{2})}{(\sqrt[3]{3}-\sqrt[3]{2})}=\\\\\\(\sqrt[3]{3}+\sqrt[3]{2})=\boxed{3^{\frac{1}{3}}+2^{\frac{1}{3}}}$


Alternativa b).