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I have an equation the following equation in my textbook, but I don't understand how it's legal for it to be simplified this way. $${1000\over \pi\sqrt[3]{500\over \pi}^2}=2\sqrt[3]{500\over \pi}$$ I know that an equivalent equation is $${1000\over \pi{500\over \pi}^{2/3}}={1000\over \pi({500\over \pi})^{1/3}({500\over \pi})^{1/3}}$$ Looking at it I can see where things might cancel, like in one of the ${500\over \pi}^{1/3}$, the $500$ could go into $1000$ twice, and cancel out the $\pi$ in the denominator, but as far as I can remember that's not allowed.

Why is this okay?

matryoshka
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    There are no math cops out there, unless you count the self-appointed ones who don't really have the power to send you to jail. So rather than "legal", you're asking about "valid" or maybe "possible". – Lisa Jul 30 '15 at 23:13

4 Answers4

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Hint:

multiply numerator and denominator for: $$ \sqrt[3]{\dfrac{500}{\pi}} $$

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

Emilio Novati
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it is $\frac{1000}{\pi\sqrt[3]{\frac{500}{\pi}}\sqrt[3]\frac{500}{\pi}}$=$\frac{\sqrt[3]{1000^3}}{\sqrt[3]{500\pi\cdot500}}$=$\sqrt[3]{\frac{1000^3}{500^2\pi}}$=$\sqrt[3]{\frac{4000}{\pi}}$=$2\sqrt[3]{\frac{500}{\pi}}$

Dr. Sonnhard Graubner
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Rewrite: $$\frac{1000}{\pi \sqrt[3]{\frac{500}{\pi}}^2}=\frac{1000}{\pi}\times \left(\frac{500}{\pi}\right)^{-2/3}$$

And note:

$$\frac{1000}{\pi}=2\left(\frac{500}{\pi}\right)^{3/3}$$

hexaflexagonal
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Note that $1000 = \sqrt[3]{1000^3}$ In general $a = \sqrt[3]{a^3}$

$\displaystyle {1000\over \pi\sqrt[3]{500\over \pi}^2}= \sqrt[3]{\frac{1000^3}{\frac{\pi^3 500^2}{\pi^2}}}$

And you have to remember that $1000^3 = (2 \cdot 500)^3 = 8 (500)^3$

IrbidMath
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