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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 $b=0$ )