## AriPotta 2 years ago i'm really feeling like an idiot. i completely forgot how to do this. can someone refresh my memory?

1. ParthKohli

$\large b^{n/a}=\sqrt[a]{b^n}$

2. ParthKohli

Yeah, but because that's a negative, I think you should put a $$\large \sqrt[3]{216^{-2}}$$

3. ParthKohli

Remember this guy?$\large a^{-n} = \dfrac{1}{a^n}$

4. ParthKohli

Yes, and now find the cube root of that thing.

5. chihiroasleaf

use this $a ^{-n}=\frac{ 1 }{a ^{n} }$

6. agent0smith

$216^{\frac{ -2 }{ 3 }} = \frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \sqrt[3]{216^{2}}}$

7. agent0smith

Easier to think of it that way

8. 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

9. ParthKohli

An easier way to think about it is to find the cube root of the numerator and the denominator separately

10. agent0smith

$\frac{ 1 }{216^{\frac{ 2 }{ 3 }} } = \frac{ 1 }{ \left( 216^\frac{ 1 }{ 3 } \right)^2} = \frac{ 1 }{ \sqrt[3]{216^{2}}}$

11. ParthKohli

To make some stuff simpler,$\dfrac{1}{\sqrt[3]{(6^3)^2}} = \dfrac{1}{\sqrt[3]{(6^2)^3}}$

12. ParthKohli

Yeah 8-)

13. AriPotta

thanks :)

14. 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.