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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

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Is the initial expression $\sqrt[4]{\dfrac{g^3 h^4}{4 r^{14} } }$ or something else? –  Henry Sep 8 '11 at 21:49
    
Where did you get the extra $h^3$ in the numerator in your first step? $\sqrt[4]{g^3h^4}=h\sqrt[4]{g^3}$. –  Brian M. Scott Sep 8 '11 at 21:54
    
How do the series of expressions after "I did this" relate to each other? –  Henning Makholm Sep 8 '11 at 21:54
    
How did $\frac{1}{4r^{14}}$ turn into $h^3$? –  anon Sep 8 '11 at 21:55
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@anon: I edited for readability, but did not include anything extra. Seems to me the OP is first trying to simplify the numerator, $\sqrt[4]{g^3h^4}$ (alas, incorrectly, since that extra $h^3$ shouldn't be there); and then separately trying to simplify the denominator $\sqrt[4]{4r^{14}}$ (again, unfortunately incorrectly since the $4$ should not have become a $2$). –  Arturo Magidin Sep 8 '11 at 22:06
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2 Answers

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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Don't forget that "for $a$ and $b$ positive". No hypotheses on $g$, $h$ and $r$ were stated in the problem (but maybe they were in the original). –  Robert Israel Sep 8 '11 at 23:00
    
Is $r^3 \sqrt{2r}$ "simpler" than $\sqrt{2r^7}$? –  user7530 Sep 9 '11 at 0:11
    
@Robert: True; some absolute values should go there. $g$ must be positive for the expression to be sensible, but $h$ and $r$ may be negative. –  Arturo Magidin Sep 9 '11 at 0:46
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If the initial expression is $$\sqrt[4]{\dfrac{g^3 h^4}{4 r^{14} } }$$ then you have made slight errors as $\sqrt[4]{g^3 h^4} = h\sqrt[4]{g^3}$ not $h\sqrt[4]{g^3 h^3}$, while $\sqrt[4]{4 r^{14}} = r^3 \sqrt[4]{4 r^2}$ not $r^3 \sqrt[4]{2 r^2}$.

But otherwise you seem to have done sensible things.

So you could end up with $$\frac{h}{r^3}\sqrt[4]{\dfrac{g^3 }{4 r^{2} } }$$ or write it some other way, such as $$2^{-0.5} g^{0.75} h^1 r^{-3.5}.$$

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