<s>[INST] Simplifying the expression $\sqrt[4]{\frac{g^3h^4}{4r^{14}}}$? How would I simplify this expression?
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}}\ ?$$
I did this
$$\begin{align*}
\sqrt[4]{g^3h^3h^4}\\
h\sqrt[4]{g^3h^3}\\
\sqrt[4]{4r^{14}}\\
\sqrt[4]{2r^2r^{12}}\\
r^3\sqrt[4]{2r^2}\\
\end{align*}$$
But I am stuck?
Yes that is correct
 [/INST] Why did $h^4$ become $h^3h^4$? Why did $4$ become $2$?
In general, remember that for $a$ and $b$ positive, $\sqrt[4]{ab} = \sqrt[4]{a}\sqrt[4]{b}$, and that $(a^r)^{s} = a^{rs}$. Added: If $g$, $h$, and $r$ are positive, then you can rewrite what you have as:
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}} = \left( g^3\times h^4 \times 4^{-1} \times r^{-14}\right)^{1/4},$$
and then using the laws of exponents you get
$$\begin{align*}
\left(g^3\times h^4\times 4^{-1} \times r^{-14}\right)^{1/4} &= g^{3/4}\times h^{4/4} \times 4^{-1/4}\times r^{-14/4}\\
 &= g^{3/4}\times h \times 4^{-1/4}\times r^{-3}\times r^{-2/4} \\
&= \frac{h\sqrt[4]{g^3}}{r^34^{1/4}r^{1/2}}\\
&= \frac{h\sqrt[4]{g^3}}{r^3(2^2)^{1/4}r^{1/2}}\\
&=\frac{h\sqrt[4]{g^3}}{r^3 2^{1/2}r^{1/2}}\\
&= \frac{h\sqrt[4]{g^3}}{r^3\sqrt{2r}}.
\end{align*}$$
If $r$ is not known to be positive, then you shoudl replace the $r^3$ and the $r$ in the last step with $|r|^3$ and $|r|$. $g$ must be positive for the original expression to be sensible; if $h$ is not known to be positive, then you should replace the $h$ at the end with $|h|$. 
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