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$\sqrt{\sqrt{a}} = \sqrt[4]{a}$, where $a$ is a nonnegative real number

When radicals have the same index, laws of exponents like $a^m \cdot a^n = a^{m+n}$ apply to their radicands

To get rid of radicals in the denominator, multiply numerator and denominator by a radical that will give the denominator a perfect square under the radical

$\sqrt[n]{a} = a^{\frac{1}{n}}$, where $a$ is a nonnegative real number

$(\sqrt[m]{a})^n = \sqrt[m]{a^n}$, where $a$ is a nonnegative real number

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$, where $a$ and $b$ are nonnegative

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$, where $a \ge 0$ and $b > 0$