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:

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

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:

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

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

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

Now

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

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