Question

$\left[\right(64{)}^{-2}{]}^{-3}÷[{(-8{)}^2}^3{]}^2$

Answer 1

Let us consider

${\left[{\left(64\right)}^{-2}\right]}^{-3}÷{\left[{{\left(-8\right)}^2}^{{}^3}\right]}^2$

can be written as

$\frac{{\left[{\left(64\right)}^{-2}\right]}^{-3}}{{\left[{{\left(-8\right)}^2}^{{}^3}\right]}^2}$

$=\frac{{\left(64\right)}^6}{{\left(8\right)}^{12}}$ $(a^{m^n}=a^{mn})$ and power on $8$ is even so it become positive.

$=\frac{{\left(8\times 8\right)}^6}{{\left(8\right)}^{12}}$

$=\frac{8^6\times 8^6}{{\left(8\right)}^{12}}$ $(ab{)}^m=a^m{b}^m$

$=\frac{8^{12}}{8^{12}}$ $(a^m+a^m=a^{m+m})$

$=1$