1Cademy - Dividing $$\frac{p^3+q^3}{2p^2+2pq+2q^2} \div \frac{p^2-q^2}{6}$$

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How to Divide Rational Expressions

Example

Dividing $\frac{p^3+q^3}{2p^2+2pq+2q^2} \div \frac{p^2-q^2}{6}$

Divide $\frac{p^3+q^3}{2p^2+2pq+2q^2} \div \frac{p^2-q^2}{6}$ .

Step 1 — Rewrite as multiplication by the reciprocal. Flip the second fraction and change the division sign to multiplication:

$\frac{p^3+q^3}{2p^2+2pq+2q^2} \cdot \frac{6}{p^2-q^2}$

Step 2 — Factor the numerators and denominators completely. Each expression requires a different factoring technique:

$p^3 + q^3 = (p+q)(p^2 - pq + q^2)$ — sum of cubes, with $a = p$ and $b = q$ .
$2p^2 + 2pq + 2q^2 = 2(p^2 + pq + q^2)$ — factor out the GCF of $2$ .
$p^2 - q^2 = (p-q)(p+q)$ — difference of squares.
$6$ remains as $6$ .

The expression becomes:

$\frac{(p+q)(p^2-pq+q^2) \cdot 6}{2(p^2+pq+q^2)(p-q)(p+q)}$

Step 3 — Simplify by dividing out common factors. Cancel the common factor $(p+q)$ from the numerator and denominator. Also, $\frac{6}{2} = 3$ . After canceling:

$\frac{3(p^2-pq+q^2)}{(p-q)(p^2+pq+q^2)}$

Note that the trinomial factor from the sum of cubes, $p^2 - pq + q^2$ , and the trinomial remaining in the denominator, $p^2 + pq + q^2$ , differ in the sign of the middle term and therefore do not cancel. This example combines three factoring techniques within a single division problem — the sum of cubes pattern, GCF extraction, and the difference of squares pattern — demonstrating how different special product formulas work together when dividing rational expressions involving two variables.

Updated 2026-04-21

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OpenStax Elementary Algebra

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