Simplification algebraic of a cube root. I am trying to simplify this:
$$\frac{1000}{\pi \cdot (\frac{500}{\pi})^{\frac{2}{3}}}$$
and I think it becomes:
$$2 \cdot \sqrt[3]{\frac{500}{\pi}}$$
I basically thought we cube root the $\frac{500}{\pi}$ and then multiply the denominator by $\frac{500}{\pi}$ which could cancel out some stuff. This is how I thought about it:
$$\frac{1000}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot \frac{500}{\pi}}$$
$$ = \frac{1000 \cdot \pi}{\pi \cdot \sqrt[3]{\frac{500}{\pi}} \cdot 500}$$
Is there a better way to do this cancellation?
 A: I would do it like this:
$$\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{500^2}}{\sqrt[3]{\pi^2}}\cdot\dfrac{\sqrt[3]\pi}{\sqrt[3]{\pi}}}=\dfrac{1000}{\pi\cdot\frac{\sqrt[3]{250000\pi}}{\pi}}=\dfrac{1000}{\sqrt[3]{250000\pi}}=\dfrac{1000}{50\sqrt[3]{2\pi}}=\dfrac{20}{\sqrt[3]{2\pi}}\cdot\dfrac{\sqrt[3]{4\pi^2}}{\sqrt[3]{4\pi^2}}$$
$$=\dfrac{20\sqrt[3]{4\pi^2}}{2\pi}=\dfrac{10\sqrt[3]{4\pi^2}}{\pi}$$
The part where you are going wrong is where you are disregarding the square inside the cube root of $\frac {500}{\pi}$, and you are just "taking it out" of the cube root to the denominator.
This is the step where you have gone wrong:$$\dfrac{1000}{\pi\cdot \sqrt[3]{\frac{500}{\pi}}\cdot \frac {500}{\pi}}$$
A: $$\frac{1000}{\pi \cdot (\frac{500}{\pi})^{\frac{2}{3}}}=$$
$$ 2 \frac {\frac {500}{\pi}}{(\frac {500}{\pi})^\frac {2}{3}}=$$
$$ 2(\frac {500}{\pi})^\frac {1}{3}=$$
$$10(\frac {4}{\pi})^\frac {1}{3}$$
A: $$\frac{1000}{ \pi \cdot (\frac{500}{\pi})^{\frac{2}{3}} }
=\frac{1000}{\pi^{\frac{1}{3}}\cdot 2^{-\frac{2}{3}}\cdot 2^{\frac{2}{3}}\cdot 500^{\frac{2}{3}}}
=\frac{1000}{\pi^{\frac{1}{3}}\cdot 2^{-\frac{2}{3}}\cdot 1000^{\frac{2}{3}}}
=\frac{2\cdot 1000^{\frac{1}{3}}}{(2\pi)^{\frac{1}{3}}}
=\frac{20}{\sqrt[3] {2\pi} }=\frac{10\sqrt[3] {4\pi^2} }{\pi}=10\sqrt[3]{\frac{4}{\pi}}$$
