(i) \(\frac{(8^{−1} × 5^3) }{ 2^{−4}}\)
Since, \(a^{−m} = \frac{1}{a^m},a^m÷a^n = a^{m−n}\)
\(\frac{(8^{−1} × 5^3) }{ 2^{−4}}\)
\(= \frac{(2^4 × 5^3)}{8^1}\) [Since \(a^{−m} = \frac{1}{a^m}\)]
\(= \frac{(2^4 × 5^3)}{2^3}\)
\(= 2^{4 − 3} × 5^3 \) [am ÷ an = am − n]
= 2 × 125
= 250
(ii) \((5^{−1}× 2^{−1}) × 6^{−1}\)
Since,\( a^m × b^m = (ab)^m, a^{-m} = \frac{1}{a^m}\)
\((5^{−1} × 2^{−1}) × 6^{−1}\)
\(=10^{−1}× 6^{−1}\)
\(= (10 × 6)^{−1} \) \( [∵a^m × b^m = (ab)^m]\)
\(= (60)^{−1} = \frac{1}{60} \) \([∵ a^{-m} = \frac{1}{a^m}]\)