Find $ \left(n-1\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ n } $ and $ B = \color{red}{ 1 }$.

$$ \begin{aligned}\left(n-1\right)^2 = \color{blue}{n^2} -2 \cdot n \cdot 1 + \color{red}{1^2} = n^2-2n+1\end{aligned} $$

Multiply $3$ by $ \dfrac{n^2-2n+1}{3} $ to get $ n^2-2n+1$.

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

Step 2: Cancel $ \color{blue}{ 3 } $ in first and second fraction.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} 3 \cdot \frac{n^2-2n+1}{3} & \xlongequal{\text{Step 1}} \frac{3}{\color{red}{1}} \cdot \frac{n^2-2n+1}{3} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{n^2-2n+1}{\color{blue}{1}} = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 \cdot \left( n^2-2n+1 \right) }{ 1 \cdot 1 } \xlongequal{\text{Step 4}} \frac{ n^2-2n+1 }{ 1 } =n^2-2n+1 \end{aligned} $$

$$ \left( \color{blue}{n^2-2n+1}\right) \cdot n = n^3-2n^2+n $$