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

Step 1: Write $ x^2+2x+1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{-x^2-2x-1}{3} $ and $ 1 $ to get $ \dfrac{ \color{purple}{ -x^2-2x+2 } }{ 3 }$.

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

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

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