I'm trying to solve a simple high school algebra problem, I would like to know if my result is correct.
Convert the radicals into exponents, solve and then express the result as a radical
$2\sqrt[3]{2}:\sqrt{(\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{3}})}$
If I call $A$ to the numerator and $B$ to the denominator, then we have $$B=((\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{3}})^{\frac{1}{2}}$$$$=(\dfrac{1}{2}\sqrt[5]{4})^{\frac{-1}{6}}$$$$=(\dfrac{1}{2})^{\frac{-1}{6}}(2^{\frac{2}{5}})^{\frac{-1}{6}}$$$$=2^{\frac{1}{6}}2^{\frac{-1}{15}}.$$
Now, using this result,we have $$\dfrac{A}{B}=\dfrac{2.2^{\frac{1}{3}}.2^{\frac{1}{15}}}{2^{\frac{1}{6}}}$$$$=2^{1-\frac{1}{6}}.2^{\frac{1}{3}}.2^{\frac{1}{15}}$$$$=2^{\frac{37}{30}}$$$$=\sqrt[30]{2^{37}}.$$
Is my solution correct? Thanks in advance.
You could simplify the result to $2\sqrt[30]{2^7}$.