Multiply $ \dfrac{2}{3} $ by $ x^2-18x+81 $ to get $ \dfrac{ 2x^2-36x+162 }{ 3 } $.

Step 1: Write $ x^2-18x+81 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{2}{3} \cdot x^2-18x+81 & \xlongequal{\text{Step 1}} \frac{2}{3} \cdot \frac{x^2-18x+81}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot \left( x^2-18x+81 \right) }{ 3 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^2-36x+162 }{ 3 } \end{aligned} $$

Subtract $2$ from $ \dfrac{-2x^2+36x-162}{3} $ to get $ \dfrac{ \color{purple}{ -2x^2+36x-168 } }{ 3 }$.

Step 1: Write $ 2 $ 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 $\color{blue}{3}$.

$$ \begin{aligned} \frac{-2x^2+36x-162}{3} -2 & \xlongequal{\text{Step 1}} \frac{-2x^2+36x-162}{3} - \frac{2}{\color{red}{1}} = \frac{ -2x^2+36x-162 }{ 3 } - \frac{ 2 \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -2x^2+36x-162 } }{ 3 } - \frac{ \color{purple}{ 6 } }{ 3 }=\frac{ \color{purple}{ -2x^2+36x-168 } }{ 3 } \end{aligned} $$