Question

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

Answer 1

$\left(i\right)\frac{27}{64}=\frac{3\times 3\times 3}{4\times 4\times 4}={\left(\frac{3}{4}\right)}^3\therefore \frac{27}{64}$ is a perfect cube
$\left(ii\right)\frac{125}{128}=\frac{5\times 5\times 5}{2\times 2\times 2\times 2\times 2\times 2\times 2}=\frac{(5{)}^3}{(2{)}^3\times (2{)}^3\times 2}$
We see that 128 is not a perfefct cube
$\therefore \frac{125}{128}$ is not a perfect cube.
$\left(iii\right)0.001331=\frac{1331}{1000000}={\frac{11\times 11\times 11}{10\times 10\times 10\times 10\times 10\times 10=\left(\frac{11}{10\times 10}\right)}}^3={\left(\frac{11}{100}\right)}^3$
$\therefore$ It is a perfect cube.

$\left(iv\right)0.04=\frac{4}{100}=\frac{2\times 2}{10\times 10}=\frac{2}{10}\times \frac{2}{10}or0.2\times 0.2$
or 0.2 is not in triplet.