Question

Find the cube root of each of the following rational numbers:
(i) $\frac{-125}{729}$
(ii) $\frac{10648}{12167}$
(iii) $\frac{-19683}{24389}$
(iv) $\frac{686}{-3456}$
(v) $\frac{-39304}{-42875}$

Answer 1

(i)
Let us consider the following rational number:

$\frac{-125}{729}$

Now

$\sqrt[3]{\frac{-125}{729}}$
$=\frac{\sqrt[3]{-125}}{\sqrt[3]{729}}$ ($\because$ $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$)
$=\frac{-\sqrt[3]{125}}{\sqrt[3]{729}}$ ($\because$ $\sqrt[3]{-a}=-\sqrt[3]{a}$ )
$=-\frac{5}{9}$ ($\because$ $729=9\times 9\times 9\text{and}125=5\times 5\times 5$)

(ii)
Let us consider the following rational number:

$\frac{10648}{12167}$

Now

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

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:
$10648=\left\{2\times 2\times 2\right\}\times \left\{11\times 11\times 11\right\}$

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:
$12167=\left\{23\times 23\times 23\right\}$

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,

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

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

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

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

(iv)
Let us consider the following rational number:

$\frac{686}{-3456}$
Now

$\sqrt[3]{\frac{686}{-3456}}$
$=-\sqrt[3]{\frac{2\times 7^3}{2^7\times 3^3}}$ (686 and 3456 are not perfect cubes; therefore, we simplify it as $\frac{686}{3456}$ by prime factorisation.)
$=-\sqrt[3]{\frac{7^3}{2^6\times 3^3}}$
$$\begin{gathered} =\frac{-\sqrt[3]{7^3}}{\sqrt[3]{2^6\times 3^3}} \\ =\frac{-7}{\sqrt[3]{2^3\times 2^3\times 3^3}} \\ =\frac{-7}{2\times 2\times 3} \\ =\frac{-7}{12} \end{gathered}$$ ($\because$ $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$)

(v)
Let us consider the following rational number:

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

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

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:
$39304=\left\{2\times 2\times 2\right\}\times \left\{17\times 17\times 17\right\}$

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:
$42875=\left\{5\times 5\times 5\right\}\times \left\{7\times 7\times 7\right\}$

Now, taking one factor from each triple, we get:
$\sqrt[3]{42875}=5\times 7=35$

Now

$\sqrt[3]{\frac{-39304}{-42875}}$
$=\frac{\sqrt[3]{-39304}}{\sqrt[3]{-42875}}$
$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}}$
$$\begin{gathered} =\frac{-34}{-35} \\ =\frac{34}{35} \end{gathered}$$