$\frac{a^2+a}{2a-12}\ast \frac{6a+6}{2a+12}:\frac{9a^3+18a^2+9a}{a^2-36}=\frac{1}{6}$
$\frac{a(a+1)}{2(a-6)}\ast \frac{6(a+1)}{2(a+6)}:\frac{9a(a^2+2a+1)}{(a-6)(a+6)}=\frac{1}{6}$
$\frac{a(a+1)}{2(a-6)}\ast \frac{3(a+1)}{a+6}:\frac{9a(a+1{)}^2}{(a-6)(a+6)}=\frac{1}{6}$
$\frac{a(a+1)}{2(a-6)}\ast \frac{3(a+1)}{a+6}\ast \frac{(a-6)(a+6)}{9a(a+1{)}^2}=\frac{1}{6}$
$\frac{1}{2}\ast \frac{1}{1}\ast \frac{1}{3}=\frac{1}{6}$
$\frac{1}{6}=\frac{1}{6}$