Find the cube-root of 1728.

Finding prime factors of 1728:

$\sqrt[3]{1728}\\[1em] = \sqrt[3]{(2 \times 2 \times 2)\times (2 \times 2 \times 2)\times (3 \times 3 \times 3)}\\[1em] = 2 \times 2 \times 3\\[1em] = 12$

Hence, $\sqrt[3]{1728} = 12$