Question
Simplify: $\big(\frac{1}{2}\big)^2-\big(\frac{1}{4}\big)^{3^{-1}}\times2^{-3}$
✓
Answer
Given, $\Big[\big(\frac{1}{2}\big)^2-\big(\frac{1}{4}\big)^3\Big]^{-1}\times2^{-3}$$=\big(\frac{1}{4}-\frac{1}{64}\big)^{-1}\times2^{-3}$$=\big(\frac{16-1}{64}\big)^{-1}\times2^{-3}$
$=\big(\frac{15}{64}\big)^{-1}\times2^{-3}$
$=\frac{64}{15}\times\frac{1}{8}$ $\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=\frac{8}{15}$
$=\big(\frac{15}{64}\big)^{-1}\times2^{-3}$
$=\frac{64}{15}\times\frac{1}{8}$ $\big[\because\text{a}^{\text{-m}}=\frac{1}{\text{a}^{\text{m}}}\big]$
$=\frac{8}{15}$
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