how would i work this out?
$\displaystyle (4^2)$
do you know that $\displaystyle a^{-1} =\frac{1}{a} \Rightarrow a^{-2} =\left (\frac{1}{a} \right )^2=\frac{1^2}{a^2}=\frac{1}{a^2} $
$\displaystyle 4^{-2} =\left (\frac{1}{4} \right ) ^2 =\left (\frac{1^2}{4^2} \right ) =\left (\frac{1 \times 1}{4 \times 4} \right )=\left (\frac{1}{16} \right )$
$\displaystyle (4^{-2})^{-3} = \left (\frac{1}{4^2} \right )^{-3} = \left ( \frac{1}{(\frac{1}{4^2})} \right )^3=(4^2)^3=4^{(2 \times 3)}=4^6$