Subtract $ \dfrac{64}{6n-48} $ from $ n^2 $ to get $ \dfrac{ \color{purple}{ 6n^3-48n^2-64 } }{ 6n-48 }$.

Step 1: Write $ n^2 $ 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 $\color{blue}{ 6n-48 }$.

$$ \begin{aligned} n^2- \frac{64}{6n-48} & \xlongequal{\text{Step 1}} \frac{n^2}{\color{red}{1}} - \frac{64}{6n-48} = \frac{ n^2 \cdot \color{blue}{ \left( 6n-48 \right) }}{ 1 \cdot \color{blue}{ \left( 6n-48 \right) }} - \frac{ 64 }{ 6n-48 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 6n^3-48n^2 } }{ 6n-48 } - \frac{ \color{purple}{ 64 } }{ 6n-48 }=\frac{ \color{purple}{ 6n^3-48n^2-64 } }{ 6n-48 } \end{aligned} $$