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?

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:$$\dfrac{1000}{\pi\cdot \sqrt[3]{\frac{500}{\pi}}\cdot \frac {500}{\pi}}$$
