What is wrong in $(\frac{1}{8})^{-\frac{1}{3}}$?

Question

I tried this:

$$\biggl (\frac{1}{8}\biggr)^{-\frac{1}{3}}= \frac{1^{-\frac{1}{3}}}{8^{-{\frac{1}{3}}}}=\frac{-\sqrt[3]{1}}{-\sqrt[3]{8}}=\frac{-1}{-2}$$

Also, is it possible to see what I don't understand here or are there maybe several things..

Answer 1

You need to use $$a^{-x} = \frac 1{a^x} \tag1$$ and $$a^{1/n}= \sqrt[n]{a} \tag 2$$

Therefore $$\left(\frac 18 \right)^{- 1/3} = \frac{1^{-1/3}}{8^{-1/3}}=\frac{1/1^{1/3}}{1/8^{1/3}}= \frac{1/\sqrt[3]1}{1/\sqrt[3]8} = \frac{1}{1/2}=2$$

Answer 2

there are some simple algebraic errors there, for instance $$a^{-\frac{1}{b}} = \frac{1}{\sqrt[b]{a}}$$ not $$ a^{-\frac{1}{b}} = -\sqrt[b]{a} $$

then

$$ \left(\frac{1}{8}\right)^{-\frac{1}{3}} = \sqrt[3]8 = 2 $$

Answer 3

$$(1/8)^{-1/3}=((1/8)^{1/3})^{-1}=(1/2)^{-1}=2$$