Question
Answer
$$\dfrac{x^3+3}{(x+1)(x-1)}=\dfrac{x^3+3}{x^2-1}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{x^3+3}{(x+1)(x-1)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{x^3+3}{x^2-x+x-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{x^3+3}{x^2-1}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{x+1}\right) $ by each term in $ \left( x-1\right) $. $$ \left( \color{blue}{x+1}\right) \cdot \left( x-1\right) = x^2 -\cancel{x}+ \cancel{x}-1 $$ |
| ② | Simplify denominator $$ x^2 \, \color{blue}{ -\cancel{x}} \,+ \, \color{blue}{ \cancel{x}} \,-1 = x^2-1 $$ |
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Simplify Rational Expressions