Question

Calcule:
$\sqrt{3+2 \sqrt[3]{2 \sqrt{2}} }$ - $\sqrt{3-2 \sqrt[3]{2 \sqrt{2}} }$.

Answer 1

Oi :]
$\sqrt{3+2 \sqrt[3]{2 \sqrt{2} } }- \sqrt{3-2 \sqrt[3]{2 \sqrt{2} } } \\ \\ \sqrt{3+2 \sqrt[3]{2 }. \sqrt[3]{ \sqrt{2} } }- \sqrt{3-2 \sqrt[3]{2 }. \sqrt[3]{ \sqrt{2} } } \\ \\ \sqrt{3+2 \sqrt[3]{2 }. \sqrt[6]{2 } }- \sqrt{3-2 \sqrt[3]{2 }. \sqrt[6]{2 } } \\ \\ \sqrt{3+2 .2^{ \frac{1}{3} }.2^{ \frac{1}{6} } }-\sqrt{3-2 .2^{ \frac{1}{3} }.2^{ \frac{1}{6} } } \\ \\ \sqrt{3+2 .2^{ \frac{1}{2} } }-\sqrt{3+2 .2^{ \frac{1}{2} } } \\ \\ \sqrt{3+2 \sqrt{2} }- \sqrt{3-2 \sqrt{2} }$

$Expressando \ 3+2 \sqrt{2}\ em \ forma \ quadratica \\ \\ 3+2 \sqrt{2}=1+2 \sqrt{2}+( \sqrt{2} )^2=( \sqrt{2}+1 ) ^2, Entao: \\ \\ \sqrt{(\sqrt{2}+1)^2} -\sqrt{(\sqrt{2}-1)^2} \ \ \ como \sqrt{x^2}= x \\ \\ (\sqrt{2}+1) -(\sqrt{2}-1) \\ \\ \sqrt{2}+1-\sqrt{2}+1 \\ \\ 1+1=2$