$$ \left( \color{blue}{a^2-b}\right) \cdot x = a^2x-bx $$

Combine like terms:

$$ x^3-ax^2+a^2x-bx = a^2x-ax^2+x^3-bx $$

Add $a^2x-ax^2+x^3-bx$ and $ \dfrac{1}{b} $ to get $ \dfrac{ \color{purple}{ a^2bx-abx^2+bx^3-b^2x+1 } }{ b }$.

Step 1: Write $ a^2x-ax^2+x^3-bx $ 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 first fraction by $\color{blue}{ b }$.

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

Multiply $x^2+ax+b$ by $ \dfrac{a^2bx-abx^2+bx^3-b^2x+1}{b} $ to get $ \dfrac{a^3bx^2+bx^5+a^2b^2x-2ab^2x^2-b^3x+ax+x^2+b}{b} $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} x^2+ax+b \cdot \frac{a^2bx-abx^2+bx^3-b^2x+1}{b} & \xlongequal{\text{Step 1}} \frac{x^2+ax+b}{\color{red}{1}} \cdot \frac{a^2bx-abx^2+bx^3-b^2x+1}{b} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2+ax+b \right) \cdot \left( a^2bx-abx^2+bx^3-b^2x+1 \right) }{ 1 \cdot b } \xlongequal{\text{Step 3}} \frac{ \cancel{a^2bx^3} -\cancel{abx^4}+bx^5 -\cancel{b^2x^3}+x^2+a^3bx^2 -\cancel{a^2bx^3}+ \cancel{abx^4}-ab^2x^2+ax+a^2b^2x-ab^2x^2+ \cancel{b^2x^3}-b^3x+b }{ b } = \\[1ex] &= \frac{a^3bx^2+bx^5+a^2b^2x-2ab^2x^2-b^3x+ax+x^2+b}{b} \end{aligned} $$