Question

Simplify the rational expression.

$\frac{1-x^2}{x^3-1}$

Short answer

The simplified rational expression is$\frac{-\left(1+x\right)}{\left(x^2+x+1\right)}$

Step-by-step solution

Step 1. Apply the concept of simplifying rational expressions.

To simplify rational expressions, we factor both numerator and denominator and use the following property of fractions:

$\frac{AC}{BC}=\frac{A}{B}$

This allows us to cancel common factors from the numerator and denominator.

Step 2. Analyze the expression.

Given$\frac{1-x^2}{x^3-1}$

This is a rational expression.

To simplify the rational expression, we factor both numerator and denominator and use this$\frac{AC}{BC}=\frac{A}{B}$

Therefore, to simplify a rational expression, we cancel factors that are common to the numerator and denominator.

Step 3. Simplify the expression.

Consider

$\begin{array}{l}\frac{1-x^2}{x^3-1}=\frac{\left(1-x\right)\left(1+x\right)}{\left(x-1\right)\left(x^2+x+1\right)}\mathrm{..............}\left[a^2-b^2=\left(a-b\right)\left(a+b\right)and{a}^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right]\\ \Rightarrow \frac{1-x^2}{x^3-1}=\frac{-\left(x-1\right)\left(1+x\right)}{\left(x-1\right)\left(x^2+x+1\right)}\end{array}$

The common factor to the numerator and denominator is$x-1$

So we will cancel the common factor from the numerator and denominator.

The simplified expression is$\frac{-\left(x-1\right)\left(1+x\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{-\left(1+x\right)}{\left(x^2+x+1\right)}$