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Answer 1 · 2018-01-23T13:03:33

it is 1. #x^0=1, AA x!=0#

Example:

#(-2)^0=(-2)^3/(-2)^3=(-2^3)/(-2^3)=1#

Answer 2 · 2018-01-23T13:08:08

See explanation.

Any number different from zero raised to #0# equals to #1#

The proof could be as follows:

#a^0=a^{b-b}=(a^b)/(a^b)=1#

The above calculations are true for all #a!=0# and all #b# (including #b=0#)