Question

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

iii) $-\frac{19683}{24389}$

Answer 1

Step: Find the cube root of $-\frac{19683}{24389}$

Cube root of $-\frac{19683}{24389}=\sqrt[3]{3-\frac{19683}{24389}}$

$\sqrt[3]{3-\frac{19683}{24389}}=\frac{\sqrt[3]{3-19683}}{\sqrt[3]{324389}}$

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

$=\frac{-\sqrt[3]{319683}}{\sqrt[3]{324389}}$

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

$\begin{array}{cc}\text{3}& \text{19683}\\ \text{3}& \text{6561}\\ \text{3}& \text{2187}\\ \text{3}& \text{729}\\ \text{3}& \text{243}\\ \text{3}& \text{81}\\ \text{3}& \text{27}\\ \text{3}& \text{9}\\ \text{3}& \text{3}\\ \text{3}& \text{1}\end{array}$


$\begin{array}{cc}\text{29}& \text{24389}\\ \text{29}& \text{841}\\ \text{29}& \text{29}\\ & \text{1}\end{array}$


$=\frac{-\sqrt[3]{33\times 3\times 3\times 3\times 3\times 3\times 3\times 3\times 3}}{\sqrt[3]{329\times 29\times 29}}$

$=-\frac{3\times 3\times 3}{29}$

$=-\frac{27}{29}$

Hence, the cube root of $-\frac{19683}{24389}$
is $-\frac{27}{29}$