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Here is what I'm trying to simplify:

$$\frac{cV_0}{b\left(\sqrt[3]{\frac{c^2V_0}{bd}}\right)^2}$$

So $b$, $c$, $d$ are all constants and $V_0=xyz$

I'm just having a ton of trouble with the algebra here.

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$$\frac{cV_0}{b\left(\sqrt[3]{\frac{c^2V_0}{bd}}\right)^2}= \frac{cV_0}{b\left(\frac{c^2V_0}{bd}\right)^{2/3}} = \frac{cV_0}{b} \left(\frac{bd}{c^2V_0}\right)^{2/3} = \frac{cV_0b^{2/3}d^{2/3}}{c^{4/3}V_0^{2/3}b} = \frac{V_0^{1/3}d^{2/3}}{c^{1/3}b^{1/3}}= \left(\frac{V_0d^2}{cb}\right)^{1/3}$$

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Hint: expand all the exponents for each term individually, such as $(\sqrt[3]{c^2})^2=c^\frac 43$, then proceed to simplify based on the result.

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Here is what you should do

$$\eqalign{ & A = {{c{V_0}} \over {b{{\left( {\root 3 \of {{{{c^2}{V_0}} \over {bd}}} } \right)}^2}}} = c{V_0}{b^{ - 1}}{\left( {{{\left( {{c^2}{V_0}{b^{ - 1}}{d^{ - 1}}} \right)}^{{1 \over 3}}}} \right)^{ - 2}} = c{V_0}{b^{ - 1}}{\left( {{c^2}{V_0}{b^{ - 1}}{d^{ - 1}}} \right)^{ - {2 \over 3}}} \cr & \,\,\, = \left( {c{c^{ - {4 \over 3}}}} \right)\left( {{V_0}V_0^{ - {2 \over 3}}} \right)\left( {{b^{ - 1}}{b^{{2 \over 3}}}} \right){d^{{2 \over 3}}} = {c^{ - {1 \over 3}}}V_0^{{1 \over 3}}{b^{ - {1 \over 3}}}{d^{{2 \over 3}}} = {\left( {{c^{ - 1}}{V_0}{b^{ - 1}}{d^2}} \right)^{{1 \over 3}}} = {\left( {{{{V_0}{d^2}} \over {cb}}} \right)^{{1 \over 3}}} \cr & \,\,\, = \root 3 \of {{{{V_0}{d^2}} \over {cb}}} \cr} $$

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