Find the cube-roots of $\dfrac{125}{216}$

Prime factors of 125:

Prime factors of 216:

$\sqrt[3]{\dfrac{125}{216}}\\[1em] = {\dfrac{\sqrt[3]{125}}{\sqrt[3]{216}}}\\[1em] = {\dfrac{\sqrt[3]{5 \times 5 \times 5}}{\sqrt[3]{(2 \times 2 \times 2)\times(3 \times 3 \times 3)}}}\\[1em] = {\dfrac{5}{2 \times 3}}\\[1em] = \dfrac{5}{6}$

Hence, $\sqrt[3]{\dfrac{125}{216}} = {\dfrac{5}{6}}$