Multiply $ \color{blue}{18x} $ by $ \left( 4-x\right) $ $$ \color{blue}{18x} \cdot \left( 4-x\right) = 72x-18x^2 $$

Simplify $ \dfrac{4-x}{72x-18x^2} $ to $ \dfrac{1}{18x} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{-x+4}$.

$$ \begin{aligned} \frac{4-x}{72x-18x^2} & =\frac{ 1 \cdot \color{blue}{ \left( -x+4 \right) }}{ 18x \cdot \color{blue}{ \left( -x+4 \right) }} = \\[1ex] &= \frac{1}{18x} \end{aligned} $$

Multiply $24x^3$ by $ \dfrac{1}{18x} $ to get $ \dfrac{ 24x^3 }{ 18x } $.

Step 1: Write $ 24x^3 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 24x^3 \cdot \frac{1}{18x} & \xlongequal{\text{Step 1}} \frac{24x^3}{\color{red}{1}} \cdot \frac{1}{18x} \xlongequal{\text{Step 2}} \frac{ 24x^3 \cdot 1 }{ 1 \cdot 18x } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 24x^3 }{ 18x } \end{aligned} $$