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AriPotta
i'm really feeling like an idiot. i completely forgot how to do this. can someone refresh my memory?
\[\large b^{n/a}=\sqrt[a]{b^n}\]
Yeah, but because that's a negative, I think you should put a \(\large \sqrt[3]{216^{-2}}\)
Remember this guy?\[\large a^{-n} = \dfrac{1}{a^n}\]
Yes, and now find the cube root of that thing.
use this \[a ^{-n}=\frac{ 1 }{a ^{n} }\]
\[216^{\frac{ -2 }{ 3 }} = \frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \sqrt[3]{216^{2}}} \]
Easier to think of it that way
here:\[\sqrt[3]{\dfrac{a}{b}} = \dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}\]I know I am smashing you with too many properties :-P
An easier way to think about it is to find the cube root of the numerator and the denominator separately
\[\frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \left( 216^\frac{ 1 }{ 3 } \right)^2} = \frac{ 1 }{ \sqrt[3]{216^{2}}}\]
To make some stuff simpler,\[\dfrac{1}{\sqrt[3]{(6^3)^2}} = \dfrac{1}{\sqrt[3]{(6^2)^3}}\]
Probably easier to find the cube root of 216, THEN square it, than square 216, then find it's cube root... both methods yield the same answer.