Find $ \left(b-8\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ b } $ and $ B = \color{red}{ 8 }$.

$$ \begin{aligned}\left(b-8\right)^2 = \color{blue}{b^2} -2 \cdot b \cdot 8 + \color{red}{8^2} = b^2-16b+64\end{aligned} $$

Multiply $ \color{blue}{b^3} $ by $ \left( b^2-16b+64\right) $ $$ \color{blue}{b^3} \cdot \left( b^2-16b+64\right) = b^5-16b^4+64b^3 $$

Simplify $ \dfrac{b-8}{b^5-16b^4+64b^3} $ to $ \dfrac{1}{b^4-8b^3} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{b-8}$.

$$ \begin{aligned} \frac{b-8}{b^5-16b^4+64b^3} & =\frac{ 1 \cdot \color{blue}{ \left( b-8 \right) }}{ \left( b^4-8b^3 \right) \cdot \color{blue}{ \left( b-8 \right) }} = \\[1ex] &= \frac{1}{b^4-8b^3} \end{aligned} $$

Multiply $a^2b$ by $ \dfrac{1}{b^4-8b^3} $ to get $ \dfrac{ a^2b }{ b^4-8b^3 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} a^2b \cdot \frac{1}{b^4-8b^3} & \xlongequal{\text{Step 1}} \frac{a^2b}{\color{red}{1}} \cdot \frac{1}{b^4-8b^3} \xlongequal{\text{Step 2}} \frac{ a^2b \cdot 1 }{ 1 \cdot \left( b^4-8b^3 \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ a^2b }{ b^4-8b^3 } \end{aligned} $$