Question
Answer
$$\dfrac{j-3}{j^3-12j^2+27j}=\dfrac{1}{j^2-9j}$$
Explanation
| $$ \begin{aligned}\frac{j-3}{j^3-12j^2+27j}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{j^2-9j}\end{aligned} $$ | |
| ① | Simplify $ \dfrac{j-3}{j^3-12j^2+27j} $ to $ \dfrac{1}{j^2-9j} $. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{j-3}$. $$ \begin{aligned} \frac{j-3}{j^3-12j^2+27j} & =\frac{ 1 \cdot \color{blue}{ \left( j-3 \right) }}{ \left( j^2-9j \right) \cdot \color{blue}{ \left( j-3 \right) }} = \\[1ex] &= \frac{1}{j^2-9j} \end{aligned} $$ |
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Simplify Rational Expressions