Question

Which of the following are perfect cubes ?
(i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533

Answer 1

(i) $64=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}$

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

Grouping the factors in triplets of equal factors, we see tghat no factor is left
216 is a perfect cube.
(iii) $243=\underset{–}{3\times 3\times 3}\times 3\times 3$
Grouping the factors in triplets, we see that two factors $3\times 3$ are left
$\therefore$ is not a perfect cube.

(iv) $1000=\underset{–}{2\times 2\times 2}\times \underset{–}{5\times 5\times 5}$
Grouping the factors in triplets of equal factors, we see that no factor is left
$\therefore 1000$ is a perfect cube.

(v) $1728=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{3\times 3\times 3}$
Grouping the factors in triplets of the equal factors, we see that no factor is left
$\therefore 1728$ is a perfrect cube.

(vi) $3087=3\times 3\times \underset{–}{7\times 7\times 7}$
Grouping the factors in triplets of the equal factors, we see that two factor $3\times 3$ are left
$\therefore$

(vii) $4608=\underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times \underset{–}{2\times 2\times 2}\times 3\times 3$
Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left
$\therefore$ 4609 is not a perfect cube.

(viii) $106480=\underset{–}{2\times 2\times }\times 2\times 5\times \underset{–}{11\times 11\times 11}$
Grouping the factors in triplets of equal factors, we see that factors 2 , 5 are left
$\therefore 106480$ is not a perfect cube.

(ix) $166375=\underset{–}{5\times 5\times 5}\times \underset{–}{11\times 11\times 11}$
Grouping the factors in triplets of equal factors, we see that no factor is left
$\therefore 166375$ is a perfect cube.

(x) $456533=\underset{–}{7\times 7\times 7}\times \underset{–}{11\times 11\times 11}$
Grouping the factors in triplets of equal factor, we see that no factor is left
$\therefore 456533$ is a perfect cube