Find the cube of $1\dfrac{2}{7}$

$\Big(1\dfrac{2}{7}\Big)^3\\[1em] = \Big(\dfrac{9}{7}\Big)^3\\[1em] = \Big(\dfrac{9}{7} \times \dfrac{9}{7} \times \dfrac{9}{7}\Big)\\[1em] = \dfrac{729}{343}\\[1em] = 2\dfrac{43}{343}$

$\Big(1\dfrac{2}{7}\Big)^3 = 2\dfrac{43}{343}$