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

(i) Let us consider the following rational number:

$\frac{-125}{729}$

Now,

$\sqrt[3]{\frac{-125}{729}}$

$=\frac{\sqrt[3]{-125}}{\sqrt[3]{729}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$

$=\frac{-\sqrt[3]{125}}{\sqrt[3]{729}} \quad(\because \sqrt[3]{-a}=-\sqrt[3]{a})$

$=-\frac{5}{9} \quad(\because 729=9 \times 9 \times 9$ and $125=5 \times 5 \times 5)$

(ii) Let us consider the following rational number:

$\frac{10648}{12167}$

Now,

$\sqrt[3]{\frac{10648}{12167}}$

$=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$

Cube root by factors:

On factorising 10648 into prime factors, we get:

$10648=2 \times 2 \times 2 \times 11 \times 11 \times 11$

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

$10648=\{2 \times 2 \times 2\} \times\{11 \times 11 \times 11\}$

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

$\sqrt[3]{10648}=2 \times 11=22$

Also

On factorising 12167 into prime factors, we get:

$12167=23 \times 23 \times 23$

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

$12167=\{23 \times 23 \times 23\}$

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

$\sqrt[3]{12167}=23$

Now

$\sqrt[3]{\frac{10648}{12167}}$

$=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}}$

$=\frac{22}{23}$

(iii) Let us consider the following rational number:

$\frac{-19683}{24389}$

Now,

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

$=\frac{\sqrt[3]{-19683}}{\sqrt[3]{24389}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$

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

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,

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

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

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

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

(iv) Let us consider the following rational number:

$\frac{686}{-3456}$

Now,

$\sqrt[3]{\frac{686}{-3456}}$

$=-\sqrt[3]{\frac{2 \times 7^{3}}{2^{7} \times 3^{3}}} \quad\left(686\right.$ and 3456 are not perfect cubes; therefore, we simplify it as $\frac{686}{3456}$ by prime factorisation.)

$=-\sqrt[3]{\frac{7^{3}}{2^{6} \times 3^{3}}}$

$=\frac{-\sqrt[3]{7^{3}}}{\sqrt[3]{2^{6} \times 3^{3}}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$

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

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

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

(v) Let us consider the following rational number:

$\frac{-39304}{-42875}$

Now,

$\sqrt[3]{\frac{-39304}{-42875}}$

$=\frac{\sqrt[3]{-39304}}{\sqrt[3]{-42875}} \quad\left(\because \sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\right)$

$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}} \quad(\because \sqrt[3]{-a}=-\sqrt[3]{a})$

Cube root by factors:

On factorising 39304 into prime factors, we get:

$39304=2 \times 2 \times 2 \times 17 \times 17 \times 17$

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

$39304=\{2 \times 2 \times 2\} \times\{17 \times 17 \times 17\}$

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

$\sqrt[3]{39304}=2 \times 17=34$

Also

On factorising 42875 into prime factors, we get:

$42875=5 \times 5 \times 5 \times 7 \times 7 \times 7$

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

$42875=\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}$

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

$\sqrt[3]{42875}=5 \times 7=35$

Now

$\sqrt[3]{\frac{-39304}{-42875}}$

$=\frac{\sqrt[3]{-39304}}{\sqrt[3]{-42875}}$

$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}}$

$=\frac{-34}{-35}$

$=\frac{34}{35}$