Question

Racionalize corretamente o denominador de$\frac{1+ \sqrt[3]{2} }{1+ \sqrt[3]{2} + \sqrt[3]{4} }$

Cálculos por favor

Answer 1

1 + ∛2 + ∛4 =
1 +∛2 + ∛2²
1 + ∛2 + ∛4 =  -  1/(1 - ∛2 )
( 1 + ∛2 ) / ( 1 + ∛2 + ∛4)
( 1 + ∛2 ) /1  *  ( 1 - ∛2)/-1
1 - ( ∛2)² /  ( -1)
(1 - ∛4) / -1
∛4 - 1  ****

Answer 2

Olá!

 Sabemos que $a^3-b^3=(a-b)(a^2+ab+b^2)$.
 
 Portanto,

$\\(1-2)=(1^{\frac{1}{3}}-2^{\frac{1}{3}})(1^{\frac{2}{3}}+1^{\frac{1}{3}}\cdot2^{\frac{1}{3}}+2^{\frac{2}{3}})\\\\ - 1 = (1 - \sqrt[3]{2})(1 + \sqrt[3]{2} + \sqrt[3]{2^2}) \\\\ 1 + \sqrt[3]{2} + \sqrt[3]{4} = - \frac{1}{1 - \sqrt[3]{2}}$

 Substituindo,

$\\ \frac{1 + \sqrt[3]{2}}{1 + \sqrt[3]{2} + \sqrt[3]{4}} = \\\\\\ (1 + \sqrt[3]{2}) \div \frac{- 1}{1 - \sqrt[3]{2}} = \\\\\\ (1 + \sqrt[3]{2}) \times \frac{1 - \sqrt[3]{2}}{- 1} = \\\\\\ \frac{1 - (\sqrt[3]{2})^2}{- 1} = \\\\\\ \frac{1 - \sqrt[3]{4}}{- 1} = \\\\\\ \boxed{\boxed{\sqrt[3]{4} - 1}}$
 
 Espero ter ajudado!