Radical Rules

$$ \begin{aligned} \sqrt[n]{a}&=a^{\frac{1}{n}}\qquad(a\geq0)\\[7pt] \sqrt[2]{a}&=\sqrt{a}\\[7pt] \sqrt{a^2}&=|a|\\[7pt] \text{For } b&=a^n \text{, we have:}\quad a=\sqrt[n]{b}\qquad (\text{only for }a\geq0,~b\geq0) \end{aligned} $$

$$ \begin{alignat}{7} \sqrt[n]{a^m} &= \bigl(a^m\bigr)^{\frac{1}{n}} &&= \Bigl(a^\frac{1}{n}\Bigr)^m &&= a^{\frac{m}{n}} &&= \Bigl(\sqrt[n]{a}\Bigr)^m \\[10pt] \sqrt[m]{\sqrt[n]{a}} &= \sqrt[m]{a^{\frac{1}{n}}} &&= \Bigl(a^\frac{1}{n}\Bigr)^{\frac{1}{m}} &&= a^{\frac{1}{m \cdot n}} &&= \sqrt[m \cdot n]{a}\\[10pt] \sqrt[n]{a} \cdot \sqrt[n]{b} &= \bigl(a^{\frac{1}{n}}\bigl) \cdot \bigl(b^{\frac{1}{n}}\bigl) &&= (a \cdot b)^\frac{1}{n} &&= \sqrt[n]{a \cdot b}\\[10pt] \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} &= \dfrac{\bigl(a^{\frac{1}{n}}\bigl)}{\bigl(b^{\frac{1}{n}}\bigl)} &&= \Bigl(\dfrac{a}{b}\Bigr)^{\frac{1}{n}} &&= \sqrt[n]{\dfrac{a}{b}} &&\qquad(b>0) \end{alignat} $$