AriPotta
  • AriPotta
i'm really feeling like an idiot. i completely forgot how to do this. can someone refresh my memory?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ParthKohli
  • ParthKohli
\[\large b^{n/a}=\sqrt[a]{b^n}\]
ParthKohli
  • ParthKohli
Yeah, but because that's a negative, I think you should put a \(\large \sqrt[3]{216^{-2}}\)
ParthKohli
  • ParthKohli
Remember this guy?\[\large a^{-n} = \dfrac{1}{a^n}\]

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More answers

ParthKohli
  • ParthKohli
Yes, and now find the cube root of that thing.
chihiroasleaf
  • chihiroasleaf
use this \[a ^{-n}=\frac{ 1 }{a ^{n} }\]
agent0smith
  • agent0smith
\[216^{\frac{ -2 }{ 3 }} = \frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \sqrt[3]{216^{2}}} \]
agent0smith
  • agent0smith
Easier to think of it that way
ParthKohli
  • ParthKohli
here:\[\sqrt[3]{\dfrac{a}{b}} = \dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}\]I know I am smashing you with too many properties :-P
ParthKohli
  • ParthKohli
An easier way to think about it is to find the cube root of the numerator and the denominator separately
agent0smith
  • agent0smith
\[\frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \left( 216^\frac{ 1 }{ 3 } \right)^2} = \frac{ 1 }{ \sqrt[3]{216^{2}}}\]
ParthKohli
  • ParthKohli
To make some stuff simpler,\[\dfrac{1}{\sqrt[3]{(6^3)^2}} = \dfrac{1}{\sqrt[3]{(6^2)^3}}\]
ParthKohli
  • ParthKohli
Yeah 8-)
AriPotta
  • AriPotta
thanks :)
agent0smith
  • agent0smith
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.

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