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

Step 1: Write $ 1-3b^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \dfrac{1}{2} $ by $ a-b $ to get $ \dfrac{ a-b }{ 2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

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

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Multiply $ \dfrac{1}{2} $ by $ a-b $ to get $ \dfrac{ a-b }{ 2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{3ab^2-3b^3-a+b}{3} $ and $ \dfrac{a-b}{2} $ to get $ \dfrac{ \color{purple}{ 6ab^2-6b^3+a-b } }{ 6 }$.

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}{ 2 }$ and the second by $\color{blue}{ 3 }$.

$$ \begin{aligned} \frac{3ab^2-3b^3-a+b}{3} + \frac{a-b}{2} & = \frac{ \left( 3ab^2-3b^3-a+b \right) \cdot \color{blue}{ 2 }}{ 3 \cdot \color{blue}{ 2 }} + \frac{ \left( a-b \right) \cdot \color{blue}{ 3 }}{ 2 \cdot \color{blue}{ 3 }} = \\[1ex] &=\frac{ \color{purple}{ 6ab^2-6b^3-2a+2b } }{ 6 } + \frac{ \color{purple}{ 3a-3b } }{ 6 }=\frac{ \color{purple}{ 6ab^2-6b^3+a-b } }{ 6 } \end{aligned} $$