Calculating $(-8)^{2/3}$ $(-8)^{2/3}$
I see this as $(-8^{1/3})^2$ = $-2^2$ =  4
Or $(-8^2)^{1/3} = 64^{1/3} = 4$ 
However, Google, Sagemath and some web calculators are saying this is equal to $-4$.
And when written as $(-8)^{2/3}$, I either get an error or $$-2 + 3.46410162 i$$
Where am I going wrong? 
 A: You were wrong when you said "I see this as..."
We, and by we I mean me and the calculators, see this as
$$-8^{2/3} = -(8^{2/3}) = -(2^2) = -(4) = -4$$
Parentheses count my friend. Be careful!

Post edit:
If you are trying to calculate $(-8)^{2/3}$, all I have to say is
welcome to the world of complex numbers! :)
A: $$-(8)^{2/3} = -\sqrt[3]{8^2} = -\sqrt[3]{64} = -4$$
Since
$$4\cdot 4\cdot 4 = 64$$
If instead the minus is inside the brakets
$$(-8)^{2/3} = 4\cdot \sqrt[3]{-1}$$
Then you can see
$$-1 = e^{i\pi}$$
And so on..
A: Note that the question was edited as this post was being made. The original question was $-8^{2/3}$.
When people write $-a^b$, they mean the negative to be outside of the exponent, so $-a^b = -(a^b)$. This is different from $(-a)^b$.
Hence,
$-8^{2/3} = -(8^{2/3}) = -4$.
Another way to remember this is via the order of operations: PEMDAS. $-$ is the same as multiplying by $-1$. Exponents come first. So do $8^{2/3}$ before applying the negative.
A: $$-8^{2/3}=-\left(\sqrt[3]{8}\right)^2=-2^2=-4$$
