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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
EdSewersPark
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### Simplifying (Sick of typing that word): cbrt[(x^0 y^4)/(3z^3

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

Should I do something about the cube roots first?

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

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

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

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

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

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

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 $3xz\sqrt[3]{3x^3z^2}$?