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

Solution:

(i)
We have:

$0.001728=\frac{1728}{1000000}$

$\therefore \sqrt[3]{0.001728}=\sqrt[3]{\frac{1728}{1000000}}=\frac{\sqrt[3]{1728}}{\sqrt[3]{1000000}}$

Now

On factorising 1728 into prime factors, we get:

$1728=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$

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

$1728=\{2 \times 2 \times 2\} \times\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\}$

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

$\sqrt[3]{1728}=2 \times 2 \times 3=12$

Also

$\sqrt[3]{1000000}=\sqrt[3]{100 \times 100 \times 100}=100$

$\therefore \sqrt[3]{0.001728}=\frac{\sqrt[3]{1728}}{\sqrt[3]{1000000}}=\frac{12}{100}=0.12$

(ii)
We have:

$0.003375=\frac{3375}{1000000}$

$\therefore \sqrt[3]{0.003375}=\sqrt[3]{\frac{3375}{1000000}}=\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}$

Now,

On factorising 3375 into prime factors, we get:

$3375=\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\}$

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

$\sqrt[3]{3375}=3 \times 5=15$

Also

$\sqrt[3]{1000000}=\sqrt[3]{100 \times 100 \times 100}=100$

$\therefore \sqrt[3]{0.003375}=\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}=\frac{15}{100}=0.15$

(iii)
We have:

$0.001=\frac{1}{1000}$

$\therefore \sqrt[3]{0.001}=\sqrt[3]{\frac{1}{1000}}=\frac{\sqrt[3]{1}}{\sqrt[3]{1000}}=\frac{1}{10}=0.1$

(iv)
We have:

$1.331=\frac{1331}{1000}$

$\therefore \sqrt[3]{1.331}=\sqrt[3]{\frac{1331}{1000}}=\frac{\sqrt[3]{1331}}{\sqrt[3]{1000}}=\frac{\sqrt[3]{11 \times 11 \times 11}}{\sqrt[3]{1000}}=\frac{11}{10}=1.1$