Question

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

Answer 1

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

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

\[= \frac{- 27}{29}\]