Multiply $ \dfrac{x^6-y^6}{x^4+x^2y^2+y^4} $ by $ x^2-2xy+y^2 $ to get $ \dfrac{ x^8-2x^7y+x^6y^2-x^2y^6+2xy^7-y^8 }{ x^4+x^2y^2+y^4 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{x^6-y^6}{x^4+x^2y^2+y^4} \cdot x^2-2xy+y^2 & \xlongequal{\text{Step 1}} \frac{x^6-y^6}{x^4+x^2y^2+y^4} \cdot \frac{x^2-2xy+y^2}{\color{red}{1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^6-y^6 \right) \cdot \left( x^2-2xy+y^2 \right) }{ \left( x^4+x^2y^2+y^4 \right) \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ x^8-2x^7y+x^6y^2-x^2y^6+2xy^7-y^8 }{ x^4+x^2y^2+y^4 } \end{aligned} $$