Question

Simplify: ${\left({64}^{\frac{2}{3}}\times 2^{-2}\times 8^0\right)}^{\frac{1}{2}}$

Answer 1

Simplify the given expression:

$\begin{array}{l}{\left({64}^{\frac{2}{3}}\times 2^{-2}\times 8^0\right)}^{\frac{1}{2}}\\ ={\left({\left(4^3\right)}^{\frac{2}{3}}\times \frac{1}{2^2}\times 1\right)}^{\frac{1}{2}}\left(\because a^0=1\mathrm{and}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right)\\ ={\left(\left(4^2\right)\times \frac{1}{2^2}\times 1\right)}^{\frac{1}{2}}\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ ={\left(16\times \frac{1}{4}\right)}^{\frac{1}{2}}\\ =\sqrt{4}\\ =2\end{array}$

The value of the given expression is ${\left({64}^{\frac{2}{3}}\times 2^{-2}\times 8^0\right)}^{\frac{1}{2}}=2$