Question

Find the cube root of
(1) $4\frac{12}{125}$ (2) $\frac{216}{2197}$

Answer 1

(1) $4\frac{12}{125}=\frac{512}{125}$


$=\frac{(2\times 2\times 2)\times (2\times 2\times 2)\times (2\times 2\times 2)}{5\times 5\times 5}\therefore \sqrt[3]{34\frac{12}{125}}=\sqrt[3]{3\frac{512}{125}}=\frac{\sqrt[3]{3512}}{\sqrt[3]{3125}}=\frac{2\times 2\times 2}{5}=\frac{8}{5}$

(2) $\frac{216}{2197}=\frac{6\times 6\times 6}{13\times 13\times 13}={\left(\frac{6}{13}\right)}^3\Rightarrow \sqrt[3]{3216}=\sqrt[3]{36\times 6\times 6}=6And,\sqrt[3]{32197}=\sqrt[3]{313\times 13\times 13}=13.\therefore \frac{\sqrt[3]{3216}}{\sqrt[3]{32197}}=\frac{6}{13}.Hence,\sqrt[3]{3\frac{216}{2197}}=\frac{\sqrt[3]{3216}}{\sqrt[3]{32197}}=\frac{6}{13}$