Multiply $7$ by $ \dfrac{n}{3n} $ to get $ \dfrac{ 7n }{ 3n } $.

Step 1: Write $ 7 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 7 \cdot \frac{n}{3n} & \xlongequal{\text{Step 1}} \frac{7}{\color{red}{1}} \cdot \frac{n}{3n} \xlongequal{\text{Step 2}} \frac{ 7 \cdot n }{ 1 \cdot 3n } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 7n }{ 3n } \end{aligned} $$

Add $ \dfrac{n+1}{n^2+7n-8} $ and $ \dfrac{7n}{3n} $ to get $ \dfrac{ \color{purple}{ 7n^3+52n^2-53n } }{ 3n^3+21n^2-24n }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 3n }$ and the second by $\color{blue}{ n^2+7n-8 }$.

$$ \begin{aligned} \frac{n+1}{n^2+7n-8} + \frac{7n}{3n} & = \frac{ \left( n+1 \right) \cdot \color{blue}{ 3n }}{ \left( n^2+7n-8 \right) \cdot \color{blue}{ 3n }} + \frac{ 7n \cdot \color{blue}{ \left( n^2+7n-8 \right) }}{ 3n \cdot \color{blue}{ \left( n^2+7n-8 \right) }} = \\[1ex] &=\frac{ \color{purple}{ 3n^2+3n } }{ 3n^3+21n^2-24n } + \frac{ \color{purple}{ 7n^3+49n^2-56n } }{ 3n^3+21n^2-24n } = \\[1ex] &=\frac{ \color{purple}{ 7n^3+52n^2-53n } }{ 3n^3+21n^2-24n } \end{aligned} $$