Originally Posted by

**uperkurk** ok so I'm looking at this one, hopefully you can point me in the right direction.

$\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}$

I'll say I need to do something like this $\displaystyle \sqrt[4]{\frac{2a^8}{b^2c^3}}\times \sqrt[4]{\frac{27b}{27b}} = \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}}$ so now I have the denominators all the same, now what? I could take the cube root of most things?

$\displaystyle \sqrt[4]{\frac{54a^8b}{3^3b^3c^3}} = \frac{\sqrt[4]{6\cdot 9\cdot a^8\cdot b}}{\sqrt{3bc}}$ I'm lost here...

**EDIT** nah that surely isn't right, the 4th root is making it confusing