Question
Answer
$$\dfrac{a^2-b^2}{b^3}\dfrac{a^2}{b-ab}=\dfrac{a^4-a^2b^2}{-ab^4+b^4}$$
Explanation
| $$ \begin{aligned}\frac{a^2-b^2}{b^3}\frac{a^2}{b-ab}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{a^4-a^2b^2}{-ab^4+b^4}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{a^2-b^2}{b^3} $ by $ \dfrac{a^2}{b-ab} $ to get $ \dfrac{a^4-a^2b^2}{-ab^4+b^4} $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{a^2-b^2}{b^3} \cdot \frac{a^2}{b-ab} & \xlongequal{\text{Step 1}} \frac{ \left( a^2-b^2 \right) \cdot a^2 }{ b^3 \cdot \left( b-ab \right) } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ a^4-a^2b^2 }{ b^4-ab^4 } = \frac{a^4-a^2b^2}{-ab^4+b^4} \end{aligned} $$ |
This page was created using
Simplify Rational Expressions