math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1$? I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is:
$$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
The "rule" $(a^b)^c = a^{bc}$ doesn't necessarily hold when $a &lt; 0$.
Hi this is not a fallacy but this problem overlooks some facts
i.e. there will be two square roots to every number
like.
$$
\sqrt{4} = +2 \text{ as well as} -2
$$
similarly 
$$
\sqrt{(-1)^6} = +1 \text{ or } -1
$$

So I think writing $$\sqrt{(-1)^6} = 1$$in this proof is wrong

And also people who are saying that  $$(a^b)^c != a^{b*c}$$ for a &lt;0 or something like that
please calculate this in your calculator so that it proves you worng
$$10^{2.5*log10(-1)} $$