I'm assuming you meant to write 8^(-2) or [tex]8^{-2}[/tex] where the -2 is the exponent over the 8.
If my assumption is correct, then we use the rule [tex]a^{-b} = \frac{1}{a^b}[/tex]
So,
[tex]a^{-b} = \frac{1}{a^b}\\\\8^{-2} = \frac{1}{8^2}\\\\8^{-2} = \frac{1}{64}[/tex]
Answer: Choice D. 1/64
