Evalute $\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}$ Evaluate the given expression $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$ The given answer is $\dfrac{1}{4}$. My attempt:
$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$ This is as far I as I am able to reach. Thank you!
PS I don't see how one can get $\dfrac14$. For that we have to get something like $\sqrt[n]{A^n}$.

Using the identity $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, we find that
$$
\begin{align*}
\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}} &= \sqrt[n]{\frac1{2^n}}\cdot\sqrt[n]{\frac{20}{2^4+2^2}}\\
&= \sqrt[n]{\frac1{4^n}}\cdot\sqrt[n]{\frac{20}{16+4}}\\
&=\sqrt[n]{\left(\frac14\right)^n}\cdot\sqrt[n]{\frac{20}{20}}\\
&=\frac14.
\end{align*}
$$
