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

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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

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

Step 1: Write $ x $ 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 $\color{blue}{x^2}$.

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

Subtract $6$ from $ \dfrac{x^3-2x}{x^2} $ to get $ \dfrac{ \color{purple}{ x^3-6x^2-2x } }{ x^2 }$.

Step 1: Write $ 6 $ 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 second fraction by $\color{blue}{x^2}$.

$$ \begin{aligned} \frac{x^3-2x}{x^2} -6 & \xlongequal{\text{Step 1}} \frac{x^3-2x}{x^2} - \frac{6}{\color{red}{1}} = \frac{ x^3-2x }{ x^2 } - \frac{ 6 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^3-2x } }{ x^2 } - \frac{ \color{purple}{ 6x^2 } }{ x^2 }=\frac{ \color{purple}{ x^3-6x^2-2x } }{ x^2 } \end{aligned} $$