Find the cube-roots of $-\dfrac{64}{343}$

Prime factors of 64:

Prime factors of 343:

$\sqrt[3]{-\dfrac{64}{343}}\\[1em] = {-\dfrac{\sqrt[3]{64}}{\sqrt[3]{343}}}\\[1em] = {-\dfrac{\sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2)}}{\sqrt[3]{(7 \times 7 \times 7)}}}\\[1em] = -{\dfrac{2 \times 2}{7}}\\[1em] = -{\dfrac{4}{7}}$

Hence, $\sqrt[3]{-\dfrac{64}{343}} = {-\dfrac{4}{7}}$