Question
Answer
$$\dfrac{t^4-1}{t^3+t^2+t+1}=t-1$$
Explanation
| $$ \begin{aligned}\frac{t^4-1}{t^3+t^2+t+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}t-1\end{aligned} $$ | |
| ① | Simplify $ \dfrac{t^4-1}{t^3+t^2+t+1} $ to $ t-1$. Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{t^3+t^2+t+1}$. $$ \begin{aligned} \frac{t^4-1}{t^3+t^2+t+1} & =\frac{ \left( t-1 \right) \cdot \color{blue}{ \left( t^3+t^2+t+1 \right) }}{ 1 \cdot \color{blue}{ \left( t^3+t^2+t+1 \right) }} = \\[1ex] &= \frac{t-1}{1} =t-1 \end{aligned} $$ |
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Simplify Rational Expressions