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I'm trying to do the problem


When you try to reduce the index for 40^4, its going to be 4/3. How does the index get reduced into 2x√5x? I understand 3 cubed of 40, but what happens to the 4/3

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It's hard to understand what you have written. Do you mean $\root3\of{40x^4/y^9}$? – Gerry Myerson Sep 17 '12 at 11:14

The problem is not clear. I'm going to assume it's, simplify $$\root3\of{{40x^4\over y^9}}$$ [$\TeX$ aside --- that tiny 3 looks very strange --- if someone knows how to edit it to look nicer, be my guest]

So, let's work on it one piece at a time. $$\root3\of{40}=\root3\of{8\times5}=\root3\of8\times\root3\of5=2\root3\of5$$ Then, $$\root3\of{x^4}=\root3\of{x^3\times x}=\root3\of{x^3}\times\root3\of x=x\root3\of x$$ Finally, $\root3\of{y^9}=y^3$. Putting it all together, we get $$\root3\of{{40x^4\over y^9}}={2x\root3\of{5x}\over y^3}$$

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