## Algebra 1

$-3$
Given : $(m^{2}n^{3})^{a}=\frac{1}{m^{6}n^{9}}$ (The empty square box is represented as a) This implies : $m^{2a}$ X $n^{3a}=m^{-6}$ X $n^{-9}$ (since $(ab)^{n} = a^{n}b^{n}$ and $a^{-1} = \frac{1}{a}$) Thus, equating the exponents on both sides of the equation to the corresponding bases, we get : $m^{2a}=m^{-6}$ and $n^{3a}=n^{-9}$ Hence, $2a=-6$ and $3a=-9$ This implies $a=\frac{-6}{2}=\frac{-9}{3}=-3$ Final result : The value in the blank is -3