Multiply each term of $ \left( \color{blue}{x-1}\right) $ by each term in $ \left( x^2-x+1\right) $.

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

Combine like terms:

$$ x^3 \color{blue}{-x^2} + \color{red}{x} \color{blue}{-x^2} + \color{red}{x} -1 = x^3 \color{blue}{-2x^2} + \color{red}{2x} -1 $$

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

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{1}{2} \cdot x & \xlongequal{\text{Step 1}} \frac{1}{2} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 1 \cdot x }{ 2 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ x }{ 2 } \end{aligned} $$

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

Step 1: Write $ 2x^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.

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

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

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

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

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

Add $ \dfrac{-2x^2+3x-2}{2} $ and $ \dfrac{x^3}{3} $ to get $ \dfrac{ \color{purple}{ 2x^3-6x^2+9x-6 } }{ 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}{ 3 }$ and the second by $\color{blue}{ 2 }$.

$$ \begin{aligned} \frac{-2x^2+3x-2}{2} + \frac{x^3}{3} & = \frac{ \left( -2x^2+3x-2 \right) \cdot \color{blue}{ 3 }}{ 2 \cdot \color{blue}{ 3 }} + \frac{ x^3 \cdot \color{blue}{ 2 }}{ 3 \cdot \color{blue}{ 2 }} = \\[1ex] &=\frac{ \color{purple}{ -6x^2+9x-6 } }{ 6 } + \frac{ \color{purple}{ 2x^3 } }{ 6 }=\frac{ \color{purple}{ 2x^3-6x^2+9x-6 } }{ 6 } \end{aligned} $$