Find the cube-roots of 250.047

$\sqrt[3]{250.047}\\[1em] = \sqrt[3]{\dfrac{250047}{1000}}\\[1em] = \dfrac{\sqrt[3]{250047}}{\sqrt[3]{1000}}$

Prime factors of 250047:

$\therefore \dfrac{\sqrt[3]{250047}}{\sqrt[3]{1000}} = \dfrac{\sqrt[3]{(3 \times 3 \times 3) \times (3 \times 3 \times 3) \times(7 \times 7 \times 7)}}{\sqrt[3]{(10 \times 10 \times 10)}}\\[1em] = \dfrac{{3 \times 3 \times 7}}{{10}}\\[1em] = \dfrac{{63}}{{10}}\\[1em] = 6.3$

Hence, $\sqrt[3]{250.047} = 6.3$