Question

which expression is equivalent to sqrt 2 / ^3 sqrt 2?

Answer 1

 

$\displaystyle\\ \sqrt{2}=2^{^\frac{1}{2}}\\\\\sqrt[3]{2}=2^{^\frac{1}{3}}\\\\\frac{\sqrt{2}}{\sqrt[3]{2}}=\frac{2^{^\frac{1}{2}}}{2^{^\frac{1}{3}}}=2^{^{\frac{1}{2}-\frac{1}{3}}}=2^{^{\frac{3-2}{6}}}=2^{^{\frac{1}{6}}}=\boxed{\sqrt[\b6]{2}}$

Answer 2

Answer:

$\sqrt[6]{2}$

Step-by-step explanation:

write the equivalent expression for $\frac{\sqrt{2}}{\sqrt[3]{2}}$

to simplify it we need to rationalize the denominator

we multiply top and bottom by  $\sqrt[3]{4}$

$\frac{\sqrt{2}*\sqrt[3]{4}}{\sqrt[3]{2}*\sqrt[3]{4}}$

$\frac{\sqrt{2}*\sqrt[3]{4}}{\sqrt[3]{8}}$

$\frac{\sqrt{2}*\sqrt[3]{4}}{2}$

square root can be written as 1/2  and then cube root can be written as 1/3

$\sqrt[3]{4} =2^\frac{2}{3}$

$\frac{\sqrt{2}*\sqrt[3]{4}}{2}$

$\frac{2^\frac{1}{2}*2^\frac{2}{3}}{2}$

Now add the fractions $\frac{1}{2} + \frac{2}{3} =\frac{3}{6} + \frac{4}{6}=\frac{7}{6}$

$\frac{2^\frac{7}{6}}{2}$

$\frac{\sqrt[6]{2^7}}{2}$

$\frac{2\sqrt[6]{2}}{2}$

cancel out 2 at the top and bottom

$\sqrt[6]{2}$