Have wasted the better part of day on this

This problem has been vexing me for a couple of days now. Can you get someone to point out the fallacy in reasoning?

(-8)^(-2/3)

If you input this into a calculator it displays a "Math Error"
If you google this it shows some imaginary value with a real part and an imaginary part as the answer

Here is how I (and almost everyone whom I asked) went about solving the problem:
(-8)^(-2/3)=((-2)^3)^(-2/3)=(-2)^(3*(-2/3)=(-2)^(-2)=(-1/2)^2=1/4

But if you raise both sides to the power of -3/2, the negative sign "magically" disappears:
((-8)^(-2/3))^(-3/2)=(1/4)^(-3/2)
OR
(-8)^((-2/3)*(-3/2))=((2)^(-2))^(-3/2))
OR
-8=(2)^((-2)*(-3/2))
OR
-8=2^3
OR
-8=8

One jugaad I applied here was to multiply the core on the RHS by (-1)^2
(1/4)^(-3/2)=(((-1)^2)*((2)^(-2)))^(-3/2)=((-1)^(2*(-3/2)))*((2)^((-2)*(-3/2)))=((-1)^(-3))*((2)^(3))=(-1)*8=-8

('a^b' implies 'a raised to the power of b'. * denotes multiplication, while 'a/b' represents a fraction, with 'a' as the numerator and 'b' as the denominator)

I now realize it has gotten a bit messy with all the brackets. Write it down on a piece of paper. Can you spot the fallacy in reasoning?

The exponent laws do not hold in all cases.  For example, (x^y)^z=x^{yz} requires that x be a positive real number.  (It also holds if x is nonzero and y, z are integers.)

Thanks

Also, how does one solve the problem to arrive at the imaginary solution provided by google?

The negative fractional power is what gives you the paradox.
How do you multiply a anti-divider?

Here's the solution. Stop fucking with the 'i'maginary numbers. Reality gets distorted every time you do that shit. Why do you think Jews play with them all of the time? Hitler went to war against the "imaginaries" and he lost because he was dumb enough to bury himself alive in the icy hammer of mother Russia.
Also the Italians are the threat, not the Jews. Why else are olives only purchased in jars and cans?

http://www.wolframalpha.com/input/?i=%28-8%29^%28-2%2F3%29

>>6
There was nothing of any significance in that post whatsoever except for two parts:
Here's the solution.
http://www.wolframalpha.com/input/?i=%28-8%29^%28-2%2F3%29

>>7
You are welcome. God damned retards looking a gift horse in the fucking mouth. Do you know how expensive horses are, you slutty motherfucker? Just the upkeep alone is outrageous, you shit shoveler!

>>7
Why does the cube root of -1 equal (1 + i√3)/2 ?

>>9
There are multiple ways one could define x^y.  For example, 4^{1/2} could be either 2 or -2.  Similarly, for fixed x\neq0 and y=1/3 there are three possible definitions over the complex plane.

Mathematically it is a bit more elegant to have (-1)^{1/3}=(1+\sqrt{-3})/2 rather than -1: then one has z^{1/3}=e^{\log(z)/3} for a single branch of the logarithm.