Question

Find the cube root of each of the following Rational numbers:

iv) $\frac{686}{-3456}$

Answer 1

Step: Find the cube root of $\frac{686}{-3456}$

Cube root of $\frac{686}{-3456}=\sqrt[3]{3\frac{686}{-3456}}$

$\Rightarrow \sqrt[3]{3\frac{686}{-3456}}=\sqrt[3]{3\frac{2\times 343}{-2\times 1728}}=\sqrt[3]{3\frac{343}{-1728}}$

$=\frac{\sqrt[3]{3343}}{\sqrt[3]{3-1728}}$

$[\because \sqrt[3]{3\frac{a}{b}}=\frac{\sqrt[3]{3a}}{\sqrt[3]{3b}}]$

$=\frac{\sqrt[3]{3343}}{-\sqrt[3]{31728}}$

$[\because \sqrt[3]{3-a}=-\sqrt[3]{3a}]$

$\begin{array}{cc}\text{2}& \text{1728}\\ \text{2}& \text{864}\\ \text{2}& \text{432}\\ \text{2}& \text{216}\\ \text{2}& \text{108}\\ \text{2}& \text{54}\\ \text{3}& \text{27}\\ \text{3}& \text{9}\\ \text{3}& \text{1}\end{array}$

$=\frac{\sqrt[3]{37\times 7\times 7}}{-\sqrt[3]{32\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3}}$

$=-\frac{7}{-2\times 2\times 3}$

$=\frac{7}{-12}$

$=-\frac{7}{12}$

Hence, the cube root of $\frac{686}{-3456}$
is $-\frac{7}{12}$