Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$? Is $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[bc]{a}}$?
I believe it is since $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\frac{1}{\sqrt[b]{a}}^\frac{1}{c}=\frac{1}{\sqrt[bc]{a}}$.
Can someone please confirm my suspicions as I just came across a question that states in the answer that $((\frac{1}{a})^{\frac{1}{b}})^{\frac{1}{c}}=\sqrt[bc]{a}$ and I believe that this is a typo.
 A: For $a, b, c > 0$, yes, you are correct.  This is easy to check:  choose $a = 729 = 3^6$, $b = 1/2$, $c = 1/3$.  We must obviously have $((1/a)^{1/b})^{1/c} < 1$, but $\sqrt[6]{729} > 1$.
A: ${ \left( {\frac{1}{a}}^{\frac{1}{b}} \right)}^{\frac{1}{c}}= { \frac{1}{a}}^{ \frac{1}{b} \cdot \frac{1}{c}} =   { \frac{1}{a} }^{ \frac{1}{bc}} = \frac{1}{a^{\frac{1}{bc}}} = \frac{1}{\sqrt[bc]{a}}$
Thus, yes (assuming $a,b,c$ are all positive reals).
A: assuming you are working with positive real numbers and assuming you are using the definition that if $m \ne 1$ and $m > 0$ and $k \in \mathbb N$ the $m^{\frac 1k}$ is defined to be (and you have proven there is such) the unique one and only positive number $w$ so that $w^k = m$ then
$((\frac 1a)^{\frac 1b})^{\frac 1c}$ (assuming $a\ne 1$) is the unique
$\beta$ so that $\beta^c = (\frac 1a)^{\frac 1b}$.
And $(\frac 1a)^{\frac 1b}$ is the unique $\gamma$ so that $\gamma^b = \frac 1a$.
But then $\beta^{bc} = (\beta^c)^b = ((\frac 1a)^{\frac 1b})^b = \gamma^b = \frac 1a$.
So $beta$ is the unique value so that $\beta^{bc} = \frac 1a$ so $\beta = (\frac 1a)^{\frac 1{bc}$.
SO $((\frac 1a)^{\frac 1b})^{\frac 1c}=\beta = (\frac 1a)^{\frac 1{bc}}$
