Question: Why does (-8)^(1/3) become 1+Complex(1)* (3)^(1/2)?

Hello people in mapleprimes,

I have a question about why what is shown by maple by simplify(-8)^(1/3) is 1+ Complex(1)* (3)^(1/2)?

Solutions of x^3=-8 are -2, 1+Complex(1)*(3)^(1/2) and 1- Complex(1)*(3)^(1/2). And, as for the last one, 1- Complex(1)*(3)^(1/2), it is 

the conjugate of the second, so it might not need to be written, because of it being easily seen so.

Is it the same reason why just -2 is not shown as the result of simplify((-8)^(1/3))?

PS. I know the instruction to use surd in such a case.

the reason I asked this question is this:

I am reading Essential Maple, where

ln(z) = ln(rho*exp(Complex(1)*theta));

ln(rho*exp(Complex(1)*theta))=ln(rho)+Complex(1)*theta;

ln(rho)+Complex(1)*theta;=ln(rho)+Complex(1)*arctan(y,x);

and

z^a=exp(a*ln(z));

and

"Because of

exp(w*Pi*Complex(1)*k)=cos(2*Pi*k) + Complex(1)*sin(2*Pi*k);

and

cos(2*Pi*k) + Complex(1)*sin(2*Pi*k)=1;

we could equally well have chosen

ln_k z = ln(z) + 2*Pi*Complex(1)*k"

are written.

 Supposing these, there is a sentence that

"we choose k=0, and thus -Pi<=theta<=Pi to be the one (that for our canonical logarithm).

Every computer algebra language and numerical language follows this standard and takes the

complex logarithm to have its imaginary part in this range.

With this definition, (-8)^(1/3)=1 + Complex(1)*sqrt(3), and not -2. (the end of quotation)"

And, I can't understand the last sentence"With this definition", so I asked the above question.

I hope someone give an answer to the above question.

Thanks in advance.

taro