Find the cube-roots of $\dfrac{27}{64}$

Prime factors of 27:

Prime factors of 64:

$\sqrt[3]{\dfrac{27}{64}}\\[1em] = {\dfrac{\sqrt[3]{27}}{\sqrt[3]{64}}}\\[1em] = {\dfrac{\sqrt[3]{3 \times 3 \times 3}}{\sqrt[3]{(2 \times 2 \times 2)\times(2 \times 2 \times 2)}}}\\[1em] = {\dfrac{3}{2 \times 2}}\\[1em] = \dfrac{3}{4}$

Hence, $\sqrt[3]{\dfrac{27}{64}} = {\dfrac{3}{4}}$