Subtract $ \dfrac{15}{4x^3} $ from $ 18x $ to get $ \dfrac{ \color{purple}{ 72x^4-15 } }{ 4x^3 }$.

Step 1: Write $ 18x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 4x^3 }$.

$$ \begin{aligned} 18x- \frac{15}{4x^3} & \xlongequal{\text{Step 1}} \frac{18x}{\color{red}{1}} - \frac{15}{4x^3} = \frac{ 18x \cdot \color{blue}{ 4x^3 }}{ 1 \cdot \color{blue}{ 4x^3 }} - \frac{ 15 }{ 4x^3 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 72x^4 } }{ 4x^3 } - \frac{ \color{purple}{ 15 } }{ 4x^3 }=\frac{ \color{purple}{ 72x^4-15 } }{ 4x^3 } \end{aligned} $$