How do you solve -3^2/3^2

Working with the law of exponenets to solve this problem.  They say the answer is 3^0 but I can not figure out how to get that.
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1 Answer

(-3 )^(2/3)^2=(-3)^ (2*2/3)  = (-3)^(4/3)

NOT DEFINED

yukan oenlee take negativ numbers tu positiv integer powers

sept for speshal kase that square root defined tu be komplex numbers
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