Simplify $ \dfrac{x^3-9x}{x^2+6x+9} $ to $ \dfrac{x^2-3x}{x+3} $.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x+3}$.

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

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

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

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