Originally Posted by

**malaygoel** do you mean

$\displaystyle \frac{1}{({m^2n^3})^2}$*$\displaystyle \frac{1}{({m^{-3}}{n^2})^3}$

it is the same with negative powers

$\displaystyle (m^{-2})^2=m^{-2}*m^{-2}=m^{(-2)+(-2)}=m^{-4}$

hence your expression simplifies to

$\displaystyle \frac{1}{{m^4n^6}}$*$\displaystyle \frac{1}{{m^{-9}}{n^6}}$

$\displaystyle =\frac{1}{{m^4m^{-9}}}$*$\displaystyle \frac{1}{{n^6n^6}}$

$\displaystyle =\frac{1}{{m^{4+(-9)}}}$*$\displaystyle \frac{1}{{n^{6+6}}}$

$\displaystyle =\frac{1}{{m^{-5}}}$*$\displaystyle \frac{1}{{n^{12}}}$

$\displaystyle =\frac{m^{5}}{n^{12}}$

Keep Smiling

Malay