The Purplemath Forums

by **EdSewersPark** on Tue May 24, 2011 11:21 pm

$\sqrt[3][\frac{x^0y^4}{3z^3}\]$ $\sqrt[3][81x^9y^{-10}z^6]$

Should I do something about the cube roots first?

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by **stapel_eliz** on Wed May 25, 2011 12:51 am

You will need to combine the arguments of the radicals at some point, but that doesn't "have" to be your first step. :wink:

by **EdSewersPark** on Wed May 25, 2011 3:29 pm

Since they are both $\sqrt[3]{blah\,blah\,blah}$ can I combine them?

by **stapel_eliz** on Wed May 25, 2011 9:38 pm

EdSewersPark wrote:Since they are both $\sqrt[3]{blah\,blah\,blah}$ can I combine them?

Yep!

by **EdSewersPark** on Thu May 26, 2011 10:37 pm

$\sqrt[3]{\frac{81x^9z^6}{3z^3y^6}$

I know 81=27(3) and that 27 is a perfect cube and I think that looks like this $3\sqrt[3]{3x^9z^6}$ do I pull out the x^9 and z^6 so it's $3xz\sqrt[3]{3x^3z^2}$ ?

The Purplemath Forums

Simplifying (Sick of typing that word): cbrt[(x^0 y^4)/(3z^3

Simplifying (Sick of typing that word): cbrt[(x^0 y^4)/(3z^3

Re: Simplifying (Sick of typing that word): cbrt[(x^0 y^4)/(

Re: Simplifying (Sick of typing that word): cbrt[(x^0 y^4)/(