Rationalizing monomial denominators

Statement

Rationalize the denominator: (a) $\dfrac{15}{\sqrt{15}}$; (b) $\dfrac{14}{\sqrt[3]{49}}$.

Solution

(a) Multiply numerator and denominator by $\sqrt{15}$: $\frac{15}{\sqrt{15}} = \frac{15\,\sqrt{15}}{\sqrt{15}\cdot\sqrt{15}} = \frac{15\,\sqrt{15}}{15} = \sqrt{15}.$

(b) Since $\sqrt[3]{49} = \sqrt[3]{7^2}$, to complete the cube in the denominator multiply by $\sqrt[3]{7}$: $\frac{14}{\sqrt[3]{7^2}} = \frac{14\,\sqrt[3]{7}}{\sqrt[3]{7^2}\cdot\sqrt[3]{7}} = \frac{14\,\sqrt[3]{7}}{\sqrt[3]{7^3}} = \frac{14\,\sqrt[3]{7}}{7} = 2\,\sqrt[3]{7}.$

$\boxed{\sqrt{15}\ ;\quad 2\,\sqrt[3]{7}}$