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AriPotta Group Title

i'm really feeling like an idiot. i completely forgot how to do this. can someone refresh my memory?

  • one year ago
  • one year ago

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  1. ParthKohli Group Title
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    \[\large b^{n/a}=\sqrt[a]{b^n}\]

    • one year ago
  2. ParthKohli Group Title
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    Yeah, but because that's a negative, I think you should put a \(\large \sqrt[3]{216^{-2}}\)

    • one year ago
  3. ParthKohli Group Title
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    Remember this guy?\[\large a^{-n} = \dfrac{1}{a^n}\]

    • one year ago
  4. ParthKohli Group Title
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    Yes, and now find the cube root of that thing.

    • one year ago
  5. chihiroasleaf Group Title
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    use this \[a ^{-n}=\frac{ 1 }{a ^{n} }\]

    • one year ago
  6. agent0smith Group Title
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    \[216^{\frac{ -2 }{ 3 }} = \frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \sqrt[3]{216^{2}}} \]

    • one year ago
  7. agent0smith Group Title
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    Easier to think of it that way

    • one year ago
  8. ParthKohli Group Title
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    here:\[\sqrt[3]{\dfrac{a}{b}} = \dfrac{\sqrt[3]{a}}{\sqrt[3]{b}}\]I know I am smashing you with too many properties :-P

    • one year ago
  9. ParthKohli Group Title
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    An easier way to think about it is to find the cube root of the numerator and the denominator separately

    • one year ago
  10. agent0smith Group Title
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    \[\frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \left( 216^\frac{ 1 }{ 3 } \right)^2} = \frac{ 1 }{ \sqrt[3]{216^{2}}}\]

    • one year ago
  11. ParthKohli Group Title
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    To make some stuff simpler,\[\dfrac{1}{\sqrt[3]{(6^3)^2}} = \dfrac{1}{\sqrt[3]{(6^2)^3}}\]

    • one year ago
  12. ParthKohli Group Title
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    Yeah 8-)

    • one year ago
  13. AriPotta Group Title
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    thanks :)

    • one year ago
  14. agent0smith Group Title
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    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.

    • one year ago
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