Answer
$\dfrac{-6}{a^2}$
Work Step by Step
RECALL:
(i) $a^m \cdot a^n = a^{m+n}$
(ii) $a^{-m} = \dfrac{1}{a^m}, a \ne0$
Multiply the coefficients by each other and the variables by each other to obtain:
$=[2 \cdot (-3)] \cdot a^{5+(-7)}
\\=-6a^{-2}$
Use rule (ii) above to obtain:
$=-6 \cdot \dfrac{1}{a^2}
\\=\dfrac{-6}{a^2}$
