Multiply $ \dfrac{3x^3}{x^2-25} $ by $ \dfrac{x^2+6x+5}{x^2} $ to get $ \dfrac{ 3x^4+3x^3 }{ x^3-5x^2 } $.

Step 1: Factor numerators and denominators.

Step 2: Cancel common factors.

Step 3: Multiply numerators and denominators.

Step 4: Simplify numerator and denominator.

$$ \begin{aligned} \frac{3x^3}{x^2-25} \cdot \frac{x^2+6x+5}{x^2} & \xlongequal{\text{Step 1}} \frac{ 3x^3 }{ \left( x-5 \right) \cdot \color{red}{ \left( x+5 \right) } } \cdot \frac{ \left( x+1 \right) \cdot \color{red}{ \left( x+5 \right) } }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 3x^3 }{ x-5 } \cdot \frac{ x+1 }{ x^2 } \xlongequal{\text{Step 3}} \frac{ 3x^3 \cdot \left( x+1 \right) }{ \left( x-5 \right) \cdot x^2 } = \\[1ex] & \xlongequal{\text{Step 4}} \frac{ 3x^4+3x^3 }{ x^3-5x^2 } \end{aligned} $$