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I'm in a intermediate algebra class and am confused about how to get the simplified form of $\sqrt[3]{x^{10}}$

I tend to want to write it as $x^{10/3}$ creating a mixed fraction then simplifying that to get $x^{3}\sqrt[3]{x}$

However, when asking a friend they explained that if we look at it by going $\sqrt[3]{x^{8}}\sqrt[3]{x^{2}}$ then would get $x^2\sqrt[3]{x^{2}}$

Could someone please help with which one is correct and if the top one is correct explain why.

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Please fix the title - it is a bit misleading. – AD. May 24 '12 at 19:43
We have $x^a\cdot x^b= x^{a+b}$ and $(x^a)^b=x^{ab}$. – AD. May 24 '12 at 19:45
@AD.feel free to change the title. I also understand exponent rules but for some reason the radical is throwing me off – BandonRandon May 24 '12 at 19:47
BTW welcome to math.SE – AD. May 24 '12 at 19:56
"Simplify" is a term that cannot be defined precisely. A "simplification" that is best for one purpose is not necessarily best for another. I think that $x^{10/3}$ is a good general purpose simplification. But in a class, what is best is effectively what teacher thinks best. – André Nicolas May 24 '12 at 21:12
up vote 5 down vote accepted

$x^2\sqrt[3]{x^2}=x^2\cdot x^{2/3}=x^{8/3}\neq x^{10/3}$ ...

The mistake is at $\,\,\sqrt[3]{x^8}\sqrt[3]{x^2}\neq x^2\sqrt[3]{x^2}\,\,$ since $\,\,\sqrt[3]{x^8}\neq x^2\,\,$ but $\,x^{8/3}$

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so $x^{3}\sqrt[3]{x}$ is correct? – BandonRandon May 24 '12 at 19:51
Yes it is! $ \,\,\,\, $ – AD. May 24 '12 at 19:54

We have $$x^{10/3}=x^{(9 + 1)/3}=x^{9/3 + 1/3}=x^{3 + 1/3}=x^3x^{1/3}$$

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you can try $x^6*x^4$,we in your case it would be $x^2*x^{4/3}$,there are many forms,for example as @AD. indicated

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