Write $9^{2x+1}$ as a power of 3. Simplify $\dfrac{9a^2}{4b}\left(\dfrac{27a^3b}{6}\right)^{-1}$.

1. Express 9 as a power of 3 and simplify the exponent.

$9^{2x+1}=(3^2)^{2x+1}=3^{2(2x+1)}=3^{4x+2}$

1. Rewrite the factor with exponent $-1$ as its reciprocal.

$\dfrac{9a^2}{4b}\left(\dfrac{27a^3b}{6}\right)^{-1} =\dfrac{9a^2}{4b}\times \dfrac{6}{27a^3b}$

2. Multiply the fractions and simplify by cancelling common factors.

$\dfrac{9a^2}{4b}\times \dfrac{6}{27a^3b} =\dfrac{54a^2}{108a^3b^2} =\dfrac{1}{2}\times \dfrac{1}{a}\times \dfrac{1}{b^2} =\dfrac{1}{2ab^2}$