Question

if
$\sqrt[3]{3( \sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} } } = 2$
Then,
$\sqrt[3]{x} + \frac{1}{ \sqrt[3]{x} }$
is equal to?

Answer 1

$\sqrt[3]{3( \sqrt[3]{x} - \frac{1}{ \sqrt[3]{x}) } } = 2$

$\sqrt[3]{3} \times \sqrt[3]{ \sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} } } = 2$

${3}^{ \frac{1}{3} } \times { (\sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} } )}^{ \frac{1}{3} } = 2$

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

${( \sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} }) }^{ \frac{1}{3} \times 3} = {( \frac{2}{ {3}^{ \frac{1}{3} } } )}^{3}$

$\sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} } = \frac{ {2}^{3} }{3}$

$\sqrt[3]{x} - \frac{1}{ \sqrt[3]{x} } = \frac{8}{3}$


Rest is continued in the attachment.