Question

Evaluate $\frac{\sqrt[3]{3729}-\sqrt[3]{327}}{\sqrt[3]{3512}+\sqrt[3]{3343}}$

Answer 1

$\sqrt[3]{3729}=\sqrt[3]{39^3}=9$
$\sqrt[3]{327}=\sqrt[3]{33^3}=3$
$\sqrt[3]{3512}=\sqrt[3]{38^3}=8$
$\sqrt[3]{3343}=\sqrt[3]{37^3}=7$
$\therefore \frac{\sqrt[3]{3729}-\sqrt[3]{327}}{\sqrt[3]{3512}+\sqrt[3]{3343}}=\frac{9-3}{8+7}$
$=\frac{6}{15}=\frac{2}{5}$