Question

In the following exercises, divide the rational expressions.

$\frac{p^3+q^3}{3p^2+3pq+3q^2}÷\frac{p^2-q^2}{12}$

Short answer

The simplified rational expression is$\frac{4\left(p^2-pq+q^2\right)}{\left(p^2+pq+q^2\right)\left(p-q\right)}$

Step-by-step solution

Step 1. Given Information

We have

$\frac{p^3+q^3}{3p^2+3pq+3q^2}÷\frac{p^2-q^2}{12}=\frac{p^3+q^3}{3p^2+3pq+3q^2}\times \frac{12}{p^2-q^2}$

Step 2. Factors the numerators and denominators

$$\begin{gathered} \frac{p^3+q^3}{3p^2+3pq+3q^2}\times \frac{12}{p^2-q^2} \\ =\frac{\left(p+q\right)\left(p^2-pq+q^2\right)}{3\left(p^2+pq+q^2\right)}\times \frac{12}{\left(p-q\right)\left(p+q\right)} \\ =\frac{12\left(p+q\right)\left(p^2-pq+q^2\right)}{3\left(p^2+pq+q^2\right)\left(p-q\right)\left(p+q\right)} \end{gathered}$$

Step 3. Remove the common factors

$\frac{4\left(p^2-pq+q^2\right)}{\left(p^2+pq+q^2\right)\left(p-q\right)}$