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