Multiply $ \dfrac{25x^2-y^2}{125} $ by $ x^3 $ to get $ \dfrac{ 25x^5-x^3y^2 }{ 125 } $.

Step 1: Write $ x^3 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{25x^2-y^2}{125} \cdot x^3 & \xlongequal{\text{Step 1}} \frac{25x^2-y^2}{125} \cdot \frac{x^3}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ \left( 25x^2-y^2 \right) \cdot x^3 }{ 125 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 25x^5-x^3y^2 }{ 125 } \end{aligned} $$

Subtract $y^3$ from $ \dfrac{25x^5-x^3y^2}{125} $ to get $ \dfrac{ \color{purple}{ 25x^5-x^3y^2-125y^3 } }{ 125 }$.

Step 1: Write $ y^3 $ 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 $\color{blue}{125}$.

$$ \begin{aligned} \frac{25x^5-x^3y^2}{125} -y^3 & \xlongequal{\text{Step 1}} \frac{25x^5-x^3y^2}{125} - \frac{y^3}{\color{red}{1}} = \frac{ 25x^5-x^3y^2 }{ 125 } - \frac{ y^3 \cdot \color{blue}{ 125 }}{ 1 \cdot \color{blue}{ 125 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 25x^5-x^3y^2 } }{ 125 } - \frac{ \color{purple}{ 125y^3 } }{ 125 }=\frac{ \color{purple}{ 25x^5-x^3y^2-125y^3 } }{ 125 } \end{aligned} $$