Simplify $ \dfrac{2x-3}{8x^3-27} $ to $ \dfrac{1}{4x^2+6x+9} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{2x-3}$.

$$ \begin{aligned} \frac{2x-3}{8x^3-27} & =\frac{ 1 \cdot \color{blue}{ \left( 2x-3 \right) }}{ \left( 4x^2+6x+9 \right) \cdot \color{blue}{ \left( 2x-3 \right) }} = \\[1ex] &= \frac{1}{4x^2+6x+9} \end{aligned} $$

Multiply $ \dfrac{1}{4x^2+6x+9} $ by $ 4x^2+6x+9 $ to get $ 1$.

Step 1: Write $ 4x^2+6x+9 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Cancel $ \color{red}{ 4x^2+6x+9 } $ in first and second fraction.

Step 3: Multiply numerators and denominators.

$$ \begin{aligned} \frac{1}{4x^2+6x+9} \cdot 4x^2+6x+9 & \xlongequal{\text{Step 1}} \frac{1}{4x^2+6x+9} \cdot \frac{4x^2+6x+9}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{1}{\color{red}{1}} \cdot \frac{\color{red}{1}}{1} = \\[1ex] &= \frac{1}{1} =1 \end{aligned} $$