why $\left(\left( \left(-\frac{1}{4}\right)^{-2}\right)^\frac{1}{4}\right) \neq \left(\left(-\frac{1}{4}\right)^{-\frac{1}{2}}\right)$? Why $\left(\left( \left(-\frac{1}{4}\right)^{-2}\right)^\frac{1}{4}\right) \neq \left(\left(-\frac{1}{4}\right)^{-\frac{1}{2}}\right)$?
well, I tried this question but as far my calculations, I am getting $(((\frac {-1}{4})^{-2})^\frac {1}{4})= 2$  where as $((\frac {-1}{4})^{-1/2})= \frac {2} {i}$ .
But since $((a)^b)^c=a^{bc}$,  shouldn't these two give the same answer ?
Please help me.
Thanks in advance .
 A: Note that $\sqrt[n]{x^n}=|x|$ with $n$ even:
$$(((-4)^2)^{1/2})^{1/2}=(|-4|)^{1/2}=2$$
A: By squaring -1/4 you lose the (-1).
The real problem can be made obvious by observing the problem is roughly equal to $ ((-1)^2)^{(1/4)}=(-1)^{(1/2)}$, the specific numbers and  inverse don’t matter. Thus you will see you have $ (1)^{1/4}=(-1)^{1/2}$, something essentially  like $(1)^{1/2}=(-1)^{1/2}$, which is now obviously false.
The issue is that the powers is not always commutable, $(a^b)^c \neq (a^c)^b$.
Another fundamental issue is that on the left side you are doing calculations in real field, on the right side you are doing calculations on the complex field.  If you do both in real field and define root as solution to polynomial equation, you will see the problem more clearly, on the left side you get 2 and -2. On the right side you get nothing. If you do both in complex field, you get on the left side 2 and -2, 2i and -2i, on the right side you get 2i and -2i.
So the problem is really about how you define root and in what field you are doing your calculations. In order to understand it you probably need to learn a bit abstract algebra and complex analysis.
