Which of the following numbers are cubes of rational numbers:
(i) $ \frac{27}{64} $
(ii) $ \frac{125}{128} $
(iii) $ 0.001331 $
(iv) $ 0.04 $

To find: 

We have to find whether the given numbers are cubes of rational numbers.

Solution:

(i) $\frac{27}{64}=\frac{3 \times 3 \times 3}{4 \times 4 \times 4}$

$=\frac{3^3}{4^3}$

$=(\frac{3}{4})^3$

Therefore,

$\frac{27}{64}$ is a perfect cube.

(ii) $\frac{125}{128}=\frac{5 \times 5 \times 5}{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$=\frac{(5)^{3}}{(2)^{3} \times(2)^{3} \times 2}$

We observe that,

128 is not a perfect cube.

Therefore,

$\frac{125}{128}$ is not a perfect cube.

(iii) $0.001331=\frac{1331}{1000000}$

$= \frac{11 \times 11 \times 11}{10 \times 10 \times 10 \times 10 \times 10 \times 10}$

$=(\frac{11}{10 \times 10})^{3}$

$=(\frac{11}{100})^{3}$

Therefore,

0.001331 is a perfect cube.

(iv) $0.04=\frac{4}{100}$

$= \frac{2 \times 2}{10 \times 10}$

$=(\frac{2}{10})^{2}$

Therefore,

0.04 is not a perfect cube.

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Updated on: 10-Oct-2022

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