\(2^{-n}\)
\(2^{1 - n}\)
\(2^{n}\)
\(2^{n + 1}\)
Correct answer is A
\(8^{n} \times 2^{2n} \div 4^{3n} = 2^{3n} \times 2^{2n} \div 2^{6n}\)
\(2^{2n + 3n - 6n}\)
= \(2^{-n}\)
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