Question

Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.
(i) 125 (ii) 243 (iii) 343 (iv) 256 (v) 8000 (vi) 9261 (vii) 5324 (viii) 3375

Answer 1

(i) $125=\overline{5\times 5\times 5}=(5{)}^3$

By making them in group of 3 equal factor. We see no factor is left.
$\therefore$ It is a perfect cube.
(ii) $243=\overline{3\times 3\times 3}$

By making them in group of 3 equal factor. We see that $3\times 3$ are left
$\therefore$ It is not a perfect cube.
(iii) $343=\overline{7\times 7\times 7}$

By making them in group of 3 equal factor, we see that no factor is left.
$\therefore$ It is a perfect cube.
(iv) $256=\overline{2\times 2\times 2}\times \overline{2\times 2\times 2}\times 2\times 2$

By making them in group of 3 equal factors, we see that $2\times 2$ are left.
$\therefore$ It is not a perfect cube.
(v) 8000 = $\overline{2\times 2\times 2}\times \overline{2\times 2\times 2}\times \overline{5\times 5\times 5}$

By making them in group of 3 equal factors, we see that no factors is left.
$\therefore$ It is a perfect cube.
(vi) 9261 = $\overline{3\times 3\times 3}\times \overline{7\times 7\times 7}$

By making them in group of 3 equal factors, we see that no factor is left.
$\therefore$ It is a perfect cube.
(vii) 5324 = $2\times 2\times \overline{11\times 11\times 11}$

By making them in group of 3 equal factors, we see that $2\times 2$ are left. $\therefore$ It is not a perfect cube.
(viii) 3375 = $\overline{3\times 3\times 3}\times \overline{5\times 5\times 5}$

By grouping them in group of 3 equal factors, we see that no factors is left.
$\therefore$ It is a perfect cube.