Question

Which of the following nmbers are not perfect cubes?
(i) 64 (ii) 216 (iii) 243 (iv) 1728

Answer 1

$64=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}$

Grouping the factors in triplets, of equal factors, we see that no factor is left
$\therefore 64$ is a perfect cube.
(ii) $216=\underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}$

Grouping the factors in triplets, of equal factors, we see that no factor is left
$\therefore$ 216 is a perfect cube.
(iii) $243=\underset{–}{3\times 3\times 3}\times 3\times 3$

Grouping the factors in triplets of equal factors, we see taht $3\times 3$ are left
$\therefore$ 243 is not a perfect cube.
(iv) $1728=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}$

Grouping the factors in triplets. Of equal factors, we see that no factor is left
$\therefore$ 1728 is a perfect cube.