Answer
$$\frac{\frac{\frac{x^2-y^2}{x^2}}{x+y}}{3x}=\frac{x^2-y^2}{3x^4+3x^3y}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\frac{\frac{x^2-y^2}{x^2}}{x+y}}{3x}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\frac{x^2-y^2}{x^3+x^2y}}{3x} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^2-y^2}{3x^4+3x^3y}\end{aligned} $$ | |
| ① | Divide $ \dfrac{x^2-y^2}{x^2} $ by $ x+y $ to get $ \dfrac{ x^2-y^2 }{ x^3+x^2y } $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{x^2-y^2}{x^2} }{x+y} & \xlongequal{\text{Step 1}} \frac{x^2-y^2}{x^2} \cdot \frac{\color{blue}{1}}{\color{blue}{x+y}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2-y^2 \right) \cdot 1 }{ x^2 \cdot \left( x+y \right) } \xlongequal{\text{Step 3}} \frac{ x^2-y^2 }{ x^3+x^2y } \end{aligned} $$ |
| ② | Divide $ \dfrac{x^2-y^2}{x^3+x^2y} $ by $ 3x $ to get $ \dfrac{ x^2-y^2 }{ 3x^4+3x^3y } $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{x^2-y^2}{x^3+x^2y} }{3x} & \xlongequal{\text{Step 1}} \frac{x^2-y^2}{x^3+x^2y} \cdot \frac{\color{blue}{1}}{\color{blue}{3x}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( x^2-y^2 \right) \cdot 1 }{ \left( x^3+x^2y \right) \cdot 3x } \xlongequal{\text{Step 3}} \frac{ x^2-y^2 }{ 3x^4+3x^3y } \end{aligned} $$ |
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Simplify Rational Expressions