Find $ \left(x-1\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = x $ and $ B = 1 $.

$$ \left(x-1\right)^3 = x^3-3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2-1^3 = x^3-3x^2+3x-1 $$

Find $ \left(x-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}{ x } $ and $ B = \color{red}{ 1 }$.

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

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

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

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

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.

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

Multiply $ \dfrac{1}{6} $ by $ x-1 $ to get $ \dfrac{ x-1 }{ 6 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

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

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

Multiply $ \dfrac{1}{6} $ by $ x-1 $ to get $ \dfrac{ x-1 }{ 6 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

Add $ \dfrac{2x^3-3x^2+1}{6} $ and $ \dfrac{x-1}{6} $ to get $ \dfrac{2x^3-3x^2+x}{6} $.

To add expressions with the same denominators, we add the numerators and write the result over the common denominator.

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