# Rationalizing a Denominator with Cube Roots

Rationalize

$\dfrac{1}{\sqrt[3]{p^2}+\sqrt[3]{pq}+\sqrt[3]{q^2}}.$

How would I go about doing this without wading through lots of algebra? Is there a trick similar to how you can multiply by $\dfrac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}$ with square roots?

• A much more interesting question is rationalizing complicated denominators like $\sqrt[3]{2} + \sqrt{7}$, where tricks with simple expressions do not work anymore. – KCd Dec 21 '14 at 8:19
Hint: Multiply both numerator and denominator by $\sqrt[3]{p} - \sqrt[3]{q}$. This comes from the well-known formula: $a^3-b^3=(a-b)(a^2+ab+b^2)$