Question: 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.

Answer: 
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 $\sqrt[6]{729} > 1$.
${ \left( {\frac{1}{a}}^{\frac{1}{b}} \right)}^{\frac{1}{c}}= { \frac{1}{a}}^{ \frac{1}{b} \cdot \frac{1}{c}} = { \frac{1}{a} }^{ \frac{1}{bc}} = \frac{1}{a^{\frac{1}{bc}}} = \frac{1}{\sqrt[bc]{a}}$

Thus, yes (assuming $a,b,c$ are all positive reals).