# Find the cube root of the following rational number − 19683 24389 . - Mathematics

Sum

Find the cube root of the following rational number $\frac{- 19683}{24389}$ .

#### Solution

Let us consider the following rational number:

$\frac{- 19683}{24389}$

Now,

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

$= \frac{\sqrt[3]{- 19683}}{\sqrt[3]{24389}}$ ( ∵ $\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$ )

$= \frac{- \sqrt[3]{19683}}{\sqrt[3]{24389}}$ ( ∵ $\sqrt[3]{- a} = - \sqrt[3]{a}$ )

Cube root by factors:
On factorising 19683 into prime factors, we get:

$19683 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

On grouping the factors in triples of equal factors, we get:

$19683 = \left\{ 3 \times 3 \times 3 \right\} \times \left\{ 3 \times 3 \times 3 \right\} \times \left\{ 3 \times 3 \times 3 \right\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{19683} = 3 \times 3 \times 3 = 27$

Also
On factorising 24389 into prime factors, we get:

$24389 = 29 \times 29 \times 29$

On grouping the factors in triples of equal factors, we get:

$24389 = \left\{ 29 \times 29 \times 29 \right\}$

Now, taking one factor from each triple, we get:

$\sqrt[3]{24389} = 29$

Now
$= \frac{- \sqrt[3]{19683}}{\sqrt[3]{24389}}$
$= \frac{- 27}{29}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 4 Cubes and Cube Roots
Exercise 4.4 | Q 5.3 | Page 30