Question

Which of the following are not perfect cubes?

(i) $216$ (ii) $128$ (iii) $1000$

• A. option (ii) is correct.
• B. options (i) and (iii) are incorrect.

Answer 1

The correct option is A

option (ii) is correct.

Explanation for the correct option:

Option (ii): Determine the digit that is not a perfect cube.

Write the prime factorisation of $128$ as follows:

$128=2\times 2\times 2\times 2\times 2\times 2\times 2$

Group the factors of $128$ into triples such that there is no factor left.

$128=\left(2\times 2\times 2\right)\times \left(2\times 2\times 2\right)\times 2$

Since one factor is left, $128$ is not a perfect cube.

Hence, option (ii) is correct.

Explanation for the incorrect options:

Option (i): Determine the digits which are perfect cubes.

Write the prime factorisation of $216$ as follows:

$216=2\times 2\times 2\times 3\times 3\times 3$

Group the factors of $216$ into triples so that there is no factor left.

$216=\left(2\times 2\times 2\right)\times \left(3\times 3\times 3\right)$

Option (iii): Write the prime factorisation of $1000$ as follows:

$1000=2\times 2\times 2\times 5\times 5\times 5$

Group the factors of $1000$ into triples so that there is no factor left.

$1000=\left(2\times 2\times 2\right)\times \left(5\times 5\times 5\right)$

Therefore, options (i) and (iii) are incorrect.

Hence, option (ii) is correct.