Inactive Tutor answered 03/20/20
8^(2/6) = root6( 8*8) = root6( 2*2*2*2*2*2) = 2 = root3(8)
So in the REAL CASE,
If the fractions are equal then the expressions must also hold by substitution
Now for the imaginary case:
(-8)^(2/6) = root6[ (-1 * 2*2*2)*(-1 * 2*2*2) ] =
root6[ 2^6] = 2
If the fraction can be reduced, then there has to be common factor of 2,3,5 or some other prime.
For odd exponent factors, the negative will remain.
For even exponents, the imaginary will cancel